Let $b = 2n$ be the base for the representation. Since $n$ is a positive odd integer, $b=2n$ is an even integer and $b \ge 2$. We want to find the base-$b$ representation of $n^k$.
Let $n^k = c_m b^m + c_{m-1} b^{m-1} + \dots + c_1 b + c_0$, where $0 \le c_i < b$ are the digits.
First, let's determine the number of digits $m+1$. The number of digits is $m+1$ if $b^m \le n^k < b^{m+1}$.
Substituting $b=2n$, we have $(2n)^m \le n^k < (2n)^{m+1}$.
$2^m n^m \le n^k < 2^{m+1} n^{m+1}$.
Let's test if $m = k-1$. We need $2^{k-1} n^{k-1} \le n^k < 2^k n^k$.
The right inequality $n^k < 2^k n^k$ holds for $n \ge 1$ since $k \ge 1$.
The left inequality $2^{k-1} n^{k-1} \le n^k$ simplifies to $2^{k-1} \le n$.
So, for $n \ge 2^{k-1}$, the base-$2n$ representation of $n^k$ has exactly $k$ digits.
$n^k = c_{k-1} (2n)^{k-1} + c_{k-2} (2n)^{k-2} + \dots + c_1 (2n) + c_0$.
The digits are $c_0, c_1, \dots, c_{k-1}$. They satisfy $0 \le c_i < 2n$.
We want to show that there exists $N$ such that for all odd $n>N$, $c_i > d$ for all $i=0, 1, \dots, k-1$.

The digits $c_i$ can be computed using the standard algorithm for base conversion.
$c_0 = n^k \pmod{2n}$.
$c_1 = \lfloor n^k / (2n) \rfloor \pmod{2n}$.
$c_i = \lfloor n^k / (2n)^i \rfloor \pmod{2n}$.
$c_{k-1} = \lfloor n^k / (2n)^{k-1} \rfloor \pmod{2n}$.

Let's compute $c_0$. Since $n$ is odd, let $n=2m+1$ for some integer $m \ge 0$.
$n^2 = n(2m+1) = 2mn + n = m(2n) + n$. So $n^2 \equiv n \pmod{2n}$.
By induction, assume $n^j \equiv n \pmod{2n}$ for $j \ge 1$. Then $n^{j+1} = n^j \cdot n \equiv n \cdot n = n^2 \equiv n \pmod{2n}$.
So $n^k \equiv n \pmod{2n}$ for all $k \ge 1$.
Thus, $c_0 = n^k \pmod{2n} = n$.
We require $c_0 > d$, which means $n > d$.

Now let's compute $c_{k-1}$.
$c_{k-1} = \lfloor n^k / (2n)^{k-1} \rfloor \pmod{2n}$.
$n^k / (2n)^{k-1} = n^k / (2^{k-1} n^{k-1}) = n / 2^{k-1}$.
So $c_{k-1} = \lfloor n / 2^{k-1} \rfloor \pmod{2n}$.
We need to check if $\lfloor n / 2^{k-1} \rfloor < 2n$.
Since $k \ge 1$, $2^{k-1} \ge 1$. $n/2^{k-1} \le n$.
Also $n \ge 1$. Is $n < 2n$? Yes, for $n>0$.
So $\lfloor n/2^{k-1} \rfloor \le n/2^{k-1} \le n < 2n$.
The modulus operation is redundant. $c_{k-1} = \lfloor n / 2^{k-1} \rfloor$.
We require $c_{k-1} > d$. $\lfloor n / 2^{k-1} \rfloor > d$.
This implies $n / 2^{k-1} \ge \lfloor n / 2^{k-1} \rfloor \ge d+1$.
So $n \ge 2^{k-1}(d+1)$.

Now let's analyze the other digits $c_i$ for $1 \le i \le k-2$.
Let $Q_i = \lfloor n^k / (2n)^i \rfloor$. Then $c_i = Q_i \pmod{2n}$.
We also know that $n^k = \sum_{j=0}^{k-1} c_j (2n)^j$.
Let $R_i = n^k \pmod{(2n)^i} = \sum_{j=0}^{i-1} c_j (2n)^j$. Note $R_0=0$.
Then $n^k = Q_i (2n)^i + R_i$.
Also $R_{i+1} = n^k \pmod{(2n)^{i+1}} = \sum_{j=0}^i c_j (2n)^j$.
We have $R_{i+1} = c_i (2n)^i + R_i$.
From this, $c_i = (R_{i+1} - R_i) / (2n)^i$.
Since $0 \le R_i < (2n)^i$, we have $0 \le R_i / (2n)^i < 1$.
$R_{i+1} / (2n)^i = c_i + R_i / (2n)^i$.
Taking the floor, $\lfloor R_{i+1} / (2n)^i \rfloor = \lfloor c_i + R_i / (2n)^i \rfloor = c_i$.
So $c_i = \lfloor (n^k \pmod{(2n)^{i+1}}) / (2n)^i \rfloor$. This holds for $i=0, \dots, k-1$.

Let's find an alternative expression for $c_i$.
Let $x_i = n^k / (2n)^{i+1}$. The fractional part is ${x_i} = (n^k \pmod{(2n)^{i+1}}) / (2n)^{i+1} = R_{i+1} / (2n)^{i+1}$.
Consider $\lfloor 2n {x_i} \rfloor = \lfloor 2n R_{i+1} / (2n)^{i+1} \rfloor = \lfloor R_{i+1} / (2n)^i \rfloor = c_i$.
So $c_i = \lfloor 2n { n^k / (2n)^{i+1} } \rfloor$. This holds for $i=0, \dots, k-1$. (For $i=k-1$, $n^k / (2n)^k = 1/2^k$, ${n^k / (2n)^k} = 1/2^k$. $c_{k-1} = \lfloor 2n/2^k \rfloor = \lfloor n/2^{k-1} \rfloor$. Matches).

Let's analyze ${ n^k / (2n)^{i+1} }$.
$n^k / (2n)^{i+1} = n^{k-i-1} n^{i+1} / (2^{i+1} n^{i+1}) = n^{k-i-1} / 2^{i+1}$.
So $c_i = \lfloor 2n { n^{k-i-1} / 2^{i+1} } \rfloor$. This formula is valid for $0 \le i \le k-1$.
Let $n^{k-i-1} = q 2^{i+1} + r_i$, where $r_i = n^{k-i-1} \pmod{2^{i+1}}$, $0 \le r_i < 2^{i+1}$.
Then $n^{k-i-1} / 2^{i+1} = q + r_i / 2^{i+1}$.
Since $n$ is odd, $n^{k-i-1}$ is odd. The modulus $2^{i+1}$ is a power of 2. For $i \ge 0$, $i+1 \ge 1$, so $2^{i+1}$ is even.
An odd number cannot be a multiple of an even number greater than 1.
$n^{k-i-1}$ cannot be a multiple of $2^{i+1}$ for $i+1 \ge 1$.
So $r_i = n^{k-i-1} \pmod{2^{i+1}}$ is never 0. $r_i \ge 1$.
Also, since $n^{k-i-1}$ is odd, the remainder $r_i$ upon division by an even number must be odd. So $r_i \ge 1$.
The fractional part is ${ n^{k-i-1} / 2^{i+1} } = r_i / 2^{i+1}$.
$c_i = \lfloor 2n (r_i / 2^{i+1}) \rfloor = \lfloor n r_i / 2^i \rfloor$.
This formula holds for $i=0, \dots, k-1$.
For $i=0$, $r_0 = n^{k-1} \pmod 2$. Since $n$ is odd, $n^{k-1}$ is odd. $r_0 = 1$. $c_0 = \lfloor n(1)/2^0 \rfloor = n$. Correct.
For $i=k-1$, $r_{k-1} = n^0 \pmod{2^k} = 1$. $c_{k-1} = \lfloor n(1)/2^{k-1} \rfloor = \lfloor n/2^{k-1} \rfloor$. Correct.

We want to show $c_i > d$ for all $i=0, \dots, k-1$.
We have $c_i = \lfloor n r_i / 2^i \rfloor$. Since $r_i \ge 1$, we have $n r_i / 2^i \ge n / 2^i$.
Therefore $c_i \ge \lfloor n / 2^i \rfloor$.
To ensure $c_i > d$, it is sufficient to ensure $\lfloor n / 2^i \rfloor > d$.
This is equivalent to $\lfloor n / 2^i \rfloor \ge d+1$.
This requires $n / 2^i \ge d+1$, or $n \ge 2^i (d+1)$.

This condition must hold for all $i=0, 1, \dots, k-1$.
We need $n \ge 2^0(d+1), n \ge 2^1(d+1), \dots, n \ge 2^{k-1}(d+1)$.
The strongest condition is $n \ge 2^{k-1}(d+1)$.
Let $N = 2^{k-1}(d+1)$. Since $k \ge 1$ and $d \ge 1$, $N$ is a positive integer.
We need to show that for every odd integer $n > N$, the condition holds.
If $n > N$, then $n \ge N+1 = 2^{k-1}(d+1) + 1$.
For any $i \in {0, 1, \dots, k-1}$, we have:
$n \ge 2^{k-1}(d+1) + 1$.
$n/2^i \ge (2^{k-1}(d+1) + 1) / 2^i = 2^{k-1-i}(d+1) + 1/2^i$.
Taking the floor:
$\lfloor n/2^i \rfloor \ge \lfloor 2^{k-1-i}(d+1) + 1/2^i \rfloor$.
Since $k-1-i \ge 0$, $2^{k-1-i}$ is an integer $\ge 1$.
$2^{k-1-i}(d+1)$ is an integer.
So $\lfloor 2^{k-1-i}(d+1) + 1/2^i \rfloor = 2^{k-1-i}(d+1)$.
Thus $\lfloor n/2^i \rfloor \ge 2^{k-1-i}(d+1)$.
Since $k-1-i \ge 0$, $2^{k-1-i} \ge 1$.
So $\lfloor n/2^i \rfloor \ge 1 \cdot (d+1) = d+1$.
Since $c_i \ge \lfloor n/2^i \rfloor$, we have $c_i \ge d+1$.
Therefore, $c_i > d$.
This holds for all $i=0, 1, \dots, k-1$.
The chosen $N = 2^{k-1}(d+1)$ works.

Final check: We established earlier that $n \ge 2^{k-1}$ ensures $n^k$ has exactly $k$ digits. Since $d$ is a positive integer, $d \ge 1$. $N = 2^{k-1}(d+1) \ge 2^{k-1}(1+1) = 2^k$. The condition $n > N$ implies $n > 2^k$, which is stronger than $n \ge 2^{k-1}$. So the number of digits is indeed $k$.

The proof relies on $n$ being odd to ensure $c_0=n$ and $r_i \ge 1$. $k,d$ are positive integers. The base is $2n$.

The proof is complete.

Final Answer: The final answer is $\boxed{There exists a positive integer N such that for every odd integer n > N , the digits in the base-2n representation of n^k are all greater than d}$