$a^{2^n}-1=(a-1)(a+1)(a^2+1)...(a^{2^{n-1}}+1)$ Let’s $p|(a^{2^k} +1, a^{2^{t+k}} +1)$Suppose that prime $p$ divides $a^{2^k}+1$. Then $a^{2^k}\equiv -1$ mod $p$. So, $a^{2^{k+t}}\equiv 1$, so $p=2$. Let’s prove that $a^{2^k} +1, a^{2^{t+k}} +1$ can’t be both powers of two. $a^{2^k} +1=2^r$, where $r>1$ $(2^r-1)^{2^t}+1\equiv 2 $ mod $4$ contradiction. Therefore, $a+1,a^2+1,...,a^{2^{n-1}}+1$ have at least $n$ prime divisors, if there are power of two among them, if not we are done. If numbers above contain power of two. Then let’s consider $a-1$. If $a-1$ has odd prime divisors $q$. Then $q$ doesn’t divide $a^{2^{k}}+1$ and we get new prime divisor and we are done. So $a-1$ and $a^{2^k}+1$ are powers of two. Since $a-1>2$, then $4|a-1$, so $a^{2^k}+1\equiv 2$ mod $4$, contradiction.

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