Prove that any number consisting of $2^n$ identical digits has at least $n$ distinct prime factors.

Hint: $(10^{2^n}-1)/9$ is divisible by $10^{2^k}+1$ for $k = 0, \ldots, n-1$, and these are relatively prime.

Hint $ $ Let $\,a_{N} = 10^{\large 2^N}\!\!-1.\,$ $\,a_{N+1}\! = (10^{\large 2^N})^{\large 2}-1 = (10^{\large 2^N}\!\!+1)(10^{\large 2^N}\!\!-1) = (a_{N}\!+2) a_N $ has as at least one more prime factor than $\,a_N,\,$ because $\,1 < a_{N}\!+2\,$ is coprime to odd $a_{N}.$

Remark $\ $ This can be viewed as an analog of Euclid's proof that there are infinitely many primes when recast in the form that $\,(N+1)N\,$ has more prime factors than $N$.