Prove: $\forall$ $n\in \mathbb N, (2^n)!$ is divisible by $2^{(2^n)-1}$ and is not divisible by $2^{2^n}$ I assume induction must be used, but I'm having trouble thinking on how to use it when dealing with divisibility when there's no clear, useful way of factorizing the numbers.

No need to have a sledgehammer to crack a nut: a simple induction proof will do. We have to show $\;v_2\bigl((2^n)!\bigr)=2^n-1$.
For $n=0$, it is trivial since it means $\;v_2(1!)=2^0-1=0$.
Now suppose, by induction hypothesis, that $\;v_2\bigl((2^n)!\bigr)=2^n-1$, and let us prove that  $\;v_2\bigl((2^{n+1})!\bigr)=2^{n+1}-1$.
We'll split
$$(2^{n+1})!=(1\cdot 2\cdot 3\dotsm 2^n)\cdot(2^n+1)(2^n+2)(2^n+3)\dotsm(2^n+2^n-1)\cdot(2^n+2^n).$$
Note that, for any $k<2^n$, $\;v_2(2^n+k)=v_2(k)$, and that $\;v_2(2^n+2^n)=v_2(2\cdot2^n)=v_2(2^n)+1$, whence
\begin{align*}v_2\bigl((2^n+1)&(2^n+2)(2^n+3)\dotsm(2^n+2^n-1)\cdot(2^n+2^n)\bigr)
\\=v_2&(1\cdot 2\cdot 3\dotsm 2^n)+1=v_2\bigl((2^n)!\bigr)+1.
\end{align*}
So we have
$$v_2\bigl((2^{n+1})!\bigr)=2v_2\bigl((2^n)!\bigr)+1=2(2^n-1)+1=2^{n+1}-1.$$
