Divisors of $10^{2^{k}}+1$ proof

Question

Consider the numbers $n=1+10^{2^{k}}$ for $k=1,2,3...$ Question: Is there a proof that $n$ is not divisible by $1+10^{j}$ for all positive $j<2^{k}$?

Answer 1

Suppose for a contradiction that $(1+10^{j})|(1+10^{2^{k}})$ for some $j < 2^k$. Then we have $$(I)\ \ q(1+10^j) = 1+10^{2^k},\ q \in \mathbb{Z}$$ But notice that $q(1+10^j) = q+q\cdot10^j = 1+10^{2^{k}}$. Now notice that the expression $q+q\cdot10^j$ has last digit $q \mod 10$ while $1+10^{2^{k}}$ has last digit $1$. From here, $q$ can be $1,11,21,31...$. But when $q \ne 1$, obviously, $(I)$ cannot hold ($n$ has only first and last digit $1$ and rest $0$).

And when $q = 1$, since $j < 2^k$, we have $1+10^j < 1+10^{2^{k}}$. So we have a contradiction as required and there is no such $j$.