Prove that any number consisting of $2^n$ identical digits has at least $n$ distinct prime factors.
This is a quite well known problem. But I cant find a way to start. Any hint will be helpful.
Note: This is a problem from the book 104 Number Theory written by Titu Andrescu. There is an answer but that is not clear to me.
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 $N$.
We have declared that our number, $N$, is composed of $2^n$ identical digits,
let's call that digit $d_{0,p}$ ( let's suppose $N$ is in it's base $p$ representation.)
Therefore it must have the form:
$$N=\sum _{j=0}^{{2}^{n}-1}d_{{0,p}}\cdot{p}^{j}$$
and we know that:
$$d_{0,p} \in {\{1,2,3,...,p-1,p}\}$$
And therefore this restricts the p-adic representation of $N$ to the set:
$$ N \in \left\{ {\frac {{p}^{{2}^{n}}-1}{p-1}},2\,{\frac {{p}^{{2}^{n}}-1}{p-1}},3\,{\frac {{p}^{{ 2}^{n}}-1}{p-1}},...,{\frac {p \left( {p}^{{2}^{n}} -1 \right) }{p-1}} \right\} $$ Working on the assumption that the minimum element of such a set will have the minimum number of distinct prime factors ($\omega(N)$) of any possible $N$ for that base $p$, we find that this minimum element will always have a number of distinct prime factors equal to $n$:
$$\omega(\frac {{p}^{{2}^{n}}-1}{p-1})=n$$
