Show that $2^{n-1}$ divides $n!$ whenever $n=2^k$ for some $k \in\mathbb{N}$ 
Show that $2^{n-1}$ divides $n!$ whenever $n=2^k$ for some $k \in\mathbb{N}$

I broke down the $n!$ part into prime factors and found out that the number of '2's is $2^{k-1}$, which then shows that it is divisible by $2^{n-1}$. But I don't think this is legit? I need some help with the correct way to solve this proof, thanks guys!
 A: Hint:  Use Legendre's formula. 
A: What you did is legitimate, except that I believe your "2^(k-1)" was mean to be "$2^k - 1$" instead. However, for a more formal method, using Legendre's formula, the number of factors of $2$ in $n!$, when $n = 2^k$ for some $k \in \mathbb{N}$, is
$$\begin{equation}\begin{aligned}
v_2((2^k)!) & = \sum_{i=1}^{\infty}\left\lfloor \frac{2^k}{2^{i}} \right\rfloor \\
& = \sum_{i=0}^{k-1}2^i \\
& = 2^{k} - 1 \\
& = n - 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Thus, you have that $2^{n-1} \mid n!$, as requested to be proven.
A: Hint $:$ The power of $2$ in the prime factorization of $n!$ is given by $$\sum\limits_{i = 1}^{k} \left \lfloor \frac {n} {2^i} \right \rfloor$$ where $k = \left \lfloor \frac {\ln n} {\ln 2} \right \rfloor.$
A: If "Legendre's formula" sounds too intimidating, reason as follows.
You have to count the occurrences of the factor $2$ in $n!$ when $n=2^k$.
Every even number gives a contibution of $1$ to the total, and since every other number is even, this gives a total of $\frac122^k=2^{k-1}$ occurrences.
But every number divisble by $4=2^2$ gives an extra contribution. Since one number out of $4$ is divisible by $4$, this gives a total of $\frac142^k=2^{k-2}$ extra occurrences.
But every number divisble by $8=2^3$ gives an extra contribution. Since one number out of $8$ is divisible by $8$, this gives a total of $\frac182^k=2^{k-3}$ extra occurrences.
You repeat this very same argument $k$ times to obtain that the total number of times that $2$ appears in $n!$ is
$$
S=1+2+2^2+\cdots+2^{k-1}.
$$
It is then easy to see that $S=2^k-1$.
(Of course, this is the way to prove Legendre's formula since he argument can be easily modified to replace $2$ with any prime number)
