I couldn't find any equation for $2^k!$ so I came up with an equation that appears to work for the factorial of a power of $2$. However, I'm having problems proving it. My equation:
$$ \def\x{\times} \def\prodf#1{\prod\limits_{n=2}^{#1}\prod\limits_{i=2}^{2^{n-1}}(2i-1)} \def\ptwof#1{2^{2^#1-1}} \def\foffset{\prod_{x=2^f+1}^{2^{f+1}}} 2^k! = \ptwof{k}\x\prodf{k} $$
Induction is the obvious way to go:
$$ \begin{aligned} k=0:&\\ &2^k!=1!=1=2^{1-1}\x1=\ptwof{k}\x\prodf{k}\qquad\checkmark\\ \text{assume}&\text{true for } k=f\\ k=f+&1:\\ &2^{f+1}!=2^f!\x\foffset=\ptwof{f}\x\prodf{f}\x\foffset\\ &_{\text{unsure if this is true; did I unpack }\prodf{f}\text{ right?}}\\ &=\ptwof{f}\x(2^f+1)\x(2^f+2)\x\cdots\x(2^{f+1})\x3^{f-1}\x5^{f-2}\x7^{f-2}\x\cdots\\ &=\quad? \end{aligned} $$
I got here because
$$(2j)! = j!\x2^k\x3\x5\x7\x\cdots\x(2j-3)\x(2j-1)\\$$
So if $n=2^k$, then we can keep using that equation to replace $j!$
After trying to do, I got
$$ \ptwof{k}\x3^{p_1}\x5^{p_2}\x\cdots $$
by testing it out for some numbers and re-arranging stuff (ie I haven't proven this correct, but I guessed it).
The powers of the odd numbers were:
$$ \begin{array}{|c|c}\hline \text{power} &p &p-1 &p-1 &p-2 &p-2 &p-2 &p-2 &p-3 \\ \hline \text{number}&3 &5 &7 &9 &11 &13 &15 &17 \\ \hline \end{array}\cdots $$ Where $p=k-1$ (for $k>2$; we stop when $p-j$ is $0$)
Basically, before we move $2^i$ numbers right, the $2^i$ next numbers each get the power $p-i$, where $i$ is the number of times we moved right.
For at least up to $2^{5}!$, $\displaystyle\prodf{k}$ is the formula for the multiplying by the odd numbers.
Your formula is correct. The key point is to get the correct order for the Sylow $2$-subgroup of $S_{2^{k}},$ and you have done this. After that, the other term in your product is clearly the odd part of the order of $S_{2^{k}}.$ The proof that the order of Sylow $2$-subgroup of $S_{2^{k}}$ is what you claim can be done by induction. You can show that $S_{2^{k-1}} \wr S_{2}$ has odd index in $S_{2^{k}},$ and then everything follows. Later edit in response to comment: Here's how to get the power of $2$ right without using groups (its more or less what was done in the question): for $1 \leq j \leq 2^{k-2}-1,$ the power of $2$ dividing $2j + 2^{k-1}$ is the same as the power of $2$ dividing $2j.$ But for $j = 2^{k-2}$ the power of $2$ dividing $2j +2^{k-1}$ is $k,$ which is one greater than the power of $2$ dividing $2^{k-1}.$ Hence the power of $2$ dividing $2^{k}!$ is the same as the power of $2$ dividing $2 \times (2^{k-1}!)^{2}.$ Then by induction this power is the $(2^{k}-1)$ power.