Number of factors of summation Let $a(n)$ be the number of $1$'s in the binary expansion of $n$. If $n$ is a positive integer, show that 
$$\Bigg|\sum_{k=0}^{2^n-1}(-1)^{a(k)}\times 2^k\Bigg|$$
has at least $n!$ divisors. 
I think this can be solved using induction, but I'm not sure. The first three values of the sum are $1$, $3$, and $45$, which indeed have $1$, $2$, and $6$ factors, repectively. 

If we set
$$S(n) = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot 2^k,$$
then we note that $a(2^n + k) = a(k) + 1$, and hence
$$S(n+1) = \sum_{k=0}^{2^{n+1}-1} (-1)^{a(k)}\cdot 2^k = \sum_{k=0}^{2^n-1} (-1)^{a(k)}\cdot \left(2^k - 2^{2^n+k}\right) = -\left(2^{2^n}-1\right)\cdot S(n).$$
Thus
$$\lvert S(n)\rvert = \prod_{k=0}^{n-1} \left(2^{2^k}-1\right).$$
Now
$$\begin{align}
2^{2^n} - 1 &= \left(2^{2^{n-1}}-1\right)\left(2^{2^{n-1}}+1\right)\\
&= \underbrace{\left(2^{2^0}-1\right)}_{=1}\prod_{k=0}^{n-1} \left(2^{2^k}+1\right),
\end{align}$$
and hence
$$\lvert S(n)\rvert = \prod_{k=0}^{n-2}\left(2^{2^k}+1\right)^{n-1-k}.$$
Since the $F_k = 2^{2^k}+1$ are pairwise coprime - we have $F_n - 2= \prod\limits_{k=0}^{n-1} F_k$ by the above, so a common divisor of $F_m$ and $F_n$ for $m < n$ must divide $2 = F_n - \prod\limits_{k=0}^{n-1} F_k$, but the $F_k$ are all odd, so the $\gcd$ is $1$ -, and $F_k^m$ has at least $m+1$ divisors, the number of divisors of $\lvert S(n)\rvert$ is at least
$$\prod_{k=0}^{n-2} (n-1-k+1) = n!$$
Since $F_k$ is prime for $0 \leqslant k \leqslant 4$, we have equality for $n \leqslant 6$. $F_5 = 641 \cdot 6700417$ is composite, so for $n \geqslant 7$, $\lvert S(n)\rvert$ has strictly more divisors than $n!$. The Fermat numbers $F_k$ for $5 \leqslant k \leqslant 32$ are all known to be composite, it is an open problem whether any further Fermat primes beyond $F_4$ exist.
