The $p$-adic gamma function is defined as:
$$\Gamma_p(x)=(-1)^x\prod\limits_{\substack{k=1\\2\nmid k}}^{2^n} k.$$
It can be shown that
$$\Gamma_p(x+1)=\frac{(-1)^{x+1}(x)!}{p^{\lfloor\frac{x}{p}\rfloor}\lfloor{\frac{x}{p}}\rfloor!}.$$
I'm trying to comprehend this formula for the case $x=2^n$. I can see that $$\Gamma_p(2^n+1)=\frac{(-1)^{2^n+1}(2^n)!}{\prod\limits_{\substack{k=1\\2\mid k}}^{2^n} k},$$
but I'm having trouble seeing why $$\prod\limits_{\substack{k=1\\2\mid k}}^{2^n} k=2^{2^{n-1}}(2^{n-1})!.$$
It seems to me that
$$\prod\limits_{\substack{k=1\\2\mid k}}^{2^n} k=2\cdot4\cdot6\cdots(2^n-2)\cdot2^n=2(1\cdot2\cdot3\cdots(2^{n-1}-1)\cdot2^{n-1})=2(2^{n-1})!,$$
but this is clearly different from what I should get.
Thanks in advance to anyone who can offer some guidance.
In your step, you seem to have used the distributive property on multiplication, which is not good. $$2\cdot 4 \cdot 6 \cdot 8 = 2^4(1\cdot 2\cdot 3 \cdot 4) \neq2(1\cdot 2\cdot 3\cdot4)$$
So you should instead have $$\prod\limits_{\substack{k=1\\2\mid k}}^{2^n} k=2\cdot4\cdot6\cdots(2^n-2)\cdot2^n=2^{\color{red}{2^{n-1}}}(1\cdot2\cdot3\cdots(2^{n-1}-1)\cdot2^{n-1})=2(2^{n-1})!$$