All numbers of form $10^{k} + 1$ are composite for $k{\gt}2$ proof

Question

I have observed that all numbers of the form $10^{k} + 1$ are composite for $k{\gt}2$, and am trying to prove the same.

Progress

For odd $k$ , it is fairly obvious, as we get 11 as a divisor.

I am having difficulty with $k$ even. Have no clue how to approach. Would prefer a non-inductive approach, if possible

Developments

Thanks to crostul, we can say it is composite for $k$ having an odd divisor. Hence, it only remains to prove it for $k = 2^n$

Answer 1

The problem whether or not $G_n=10^{2^n}+1$ is composite for all $n\ge 3$ is open as well as the problem whether or not the Fermat numbers $F_n=2^{2^n}+1$ are composite for all $n\ge 5$, see here, and the article on generalized Fermat numbers.