Prove that $2^n$ does not divide $n!$

I want to prove that $2^n$ does not divide $n!$.

I was trying by induction and I'm confused about if what I'm doing is right.

First I test it with $n=1$. In fact: $$2^1 \nmid 1!$$

So if i take the I.H. as $2^n \nmid n!$ and I try to prove it for $n+1$: $$2^{n+1} \nmid (n+1)!$$ $$2^{n} \cdot 2 \nmid (n+1) \cdot n!$$

As $2^n \nmid n!$ it must be that $2^n$ divides $n+1$, so I need to prove that it doesn't. If I try by induction again I must have $n>1$ for it to be true:

$$P(n) = 2^n \nmid n+1$$ $$P(n+1) = 2^n \cdot 2 \nmid n+1+1$$ but $2^n \nmid n+1$ and $2^n \nmid 1$ because $n>1$

So my question is obviously if this is correct. I'm doubting because the exception I have to do with $n$ being greater than 1 for the second part. If I made a mistake could you point to me in a better direction? I'm sure there must be a simpler way to prove this.

Thanks!

The maximal power of $2$ which divides $n!$ is $$v_2(n!)=\lfloor \frac n2 \rfloor+\lfloor \frac n4 \rfloor +\cdots=\sum_{i=1}^{\infty} \lfloor\frac n{2^i}\rfloor$$

We bound this from above by dropping the floor function to see that $$v_2(n!)<n\sum_{i=1}^{\infty} \frac 1{2^i}=n$$

By Legendre's theorem $$\nu_2(n!) = \sum_{m\geq 1}\left\lfloor\frac{n}{2^m}\right\rfloor\leq \sum_{m\geq 1}\frac{n}{2^m}=n. $$ Equality may hold only if $2^k\parallel n$, but in such a case the $(k+1)$-th terms of the central sums fullfill a strict inequality.