Problem: Fix positive integers $k$ and $d$.
Prove that for all sufficiently large odd positive integers $n$,
the digits of the base-$2n$ representation of $n^k$ are all greater than $d$.

Solution: The problem actually doesn't have much to do with digits:
the idea is to pick any length $\ell \le k$,
and look at the rightmost $\ell$ digits of $n^k$;
that is, the remainder upon division by $(2n)^\ell$.
We compute it exactly:

\begin{claim*}
  Let $n \ge 1$ be an odd integer, and $k \ge \ell \ge 1$ integers.
  Then
  \[ n^k \bmod{(2n)^i} = c(k,\ell) \cdot n^i \]
  for some odd integer $1 \le c(k,\ell) \le 2^\ell-1$.
\end{claim*}
\begin{proof}
  This follows directly by the Chinese remainder theorem,
  with $c(k,\ell)$ being the residue class of $n^{k-i} \pmod{2^\ell}$
  (which makes sense because $n$ was odd).
\end{proof}

We can now stake the required threshold:
\begin{claim*}
  The problem statement holds once $n \ge (d+1) \cdot 2^{k-1}$.
\end{claim*}
\begin{proof}
  Suppose $n$ is that large.
  Then $n^k$ has $k$ digits in base-$2n$.
  Moreover, for each $1 \le \ell \le k$ we have
  \[ c(k,\ell) \cdot n^\ell \ge (d+1) \cdot (2n)^{\ell-1} \]
  because $n$ is large enough;
  that implies the $\ell$\ts{th} digit from the right is at least $d+1$.
  Hence the problem is solved.
\end{proof}

\begin{remark*}
  Note it doesn't really matter that $c(k,i)$ is odd \emph{per se};
  we only need that $c(k,i) \ge 1$.
\end{remark*}