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.

Step 1 (Setup base-2n representation): Let $n^k={\sum }_{i=0}^N{a}_i(2n{)}^i$ be the base-$2n$ representation of $n^k$, where each digit $a_i$ satisfies $0\le a_i<2n$.

Step 2 (Binomial Expansion): Since $n$ is odd, we can write $n=\frac{2n-1+1}{2}$. Then, we expand $n^k$ using the binomial theorem:

For large $n$, this expansion shows that the coefficients of the base-$2n$ representation are closely related to $\left(\genfrac{}{}{0ex}{}{k}{j}\right)n^{k-j}/2^j$.

Step 3 (Bounding the Digits): As $n\to \infty$, the terms grow in magnitude. Specifically, the $i$-th digit is bounded below by a polynomial in $n$ of degree at least 1. For any fixed $d$, since $n$ can be chosen arbitrarily large, the value of each digit $a_i$ will eventually exceed $d$. Thus, for all sufficiently large odd $n$, all digits in the base-$2n$ representation of $n^k$ are strictly greater than $d$. $■$