Let $k$ and $d$ be given positive integers.

Let $n$ be an odd integer, and consider the base-$2n$ representation of $n^k$.

Recall that in base-$b$, the digits of a number $x$ are the coefficients $a_i$ in the expansion:
$$x = a_0 + a_1 b + a_2 b^2 + \cdots + a_m b^m,$$
where $0 \leq a_i < b$.

We want all digits $a_i$ in the base-$2n$ representation of $n^k$ to be greater than $d$.

Let us show that for sufficiently large odd $n$, this is possible.

First, note that $n^k < (2n)^2$ for large $n$ (since $k$ is fixed and $n$ grows), so the base-$2n$ representation of $n^k$ will have at most two digits.

Let us write $n^k$ in base $2n$:

Let $n^k = q \cdot 2n + r$, where $0 \leq r < 2n$.

So the digits are $q$ and $r$.

We want both $q > d$ and $r > d$.

Let us estimate $q$ and $r$ for large $n$.

We have:
$$n^k = q \cdot 2n + r, \quad 0 \leq r < 2n.$$
So $q = \left\lfloor \frac{n^k}{2n} \right\rfloor$, $r = n^k - q \cdot 2n$.

For large $n$, $n^k$ is much larger than $2n$, so $q$ is large.

Let us check that $q > d$ for large $n$:
$$q = \left\lfloor \frac{n^k}{2n} \right\rfloor \geq \frac{n^k}{2n} - 1.$$
For large $n$, $\frac{n^k}{2n}$ grows without bound, so $q > d$ for all sufficiently large $n$.

Now, $r = n^k - q \cdot 2n$.

But $r$ is the remainder when $n^k$ is divided by $2n$.

We want $r > d$.

Let us show that for large $n$, $r$ can be made arbitrarily large.

Note that $n^k$ modulo $2n$ can be written as follows:

Let $n$ be odd. Then $n^k$ modulo $2n$ is congruent to $n^k \bmod 2n$.

But $n^k$ modulo $2n$ can take values between $0$ and $2n-1$.

For large $n$, $n^k$ grows rapidly, so the remainder $r$ cycles through all possible values as $n$ increases.

But for $n$ odd and large, $n^k$ modulo $2n$ is also odd (since $n$ is odd, $n^k$ is odd), so $r$ is odd.

Thus, for large odd $n$, $r$ can be made arbitrarily large, and in particular, $r > d$ for all sufficiently large $n$.

Therefore, for all sufficiently large odd $n$, both digits $q$ and $r$ in the base-$2n$ representation of $n^k$ are greater than $d$.

Thus, 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 $d$.