First, we prove that for any positive integer $n$, $2^n$ is not a divisor of $n!$.
Assume the opposite; then, in the prime factorization of $n!$ there should be at least $n$ factors equal to $2$. The exponent of $2$ in $n!$ is equal to $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{2^2} \rfloor + \dots + \lfloor \frac{n}{2^k} \rfloor$, where $k \in \mathbb{N}^*$, $k < n$, $\lfloor \frac{n}{2^k} \rfloor \ne 0$ and $\lfloor \frac{n}{2^{k+1}} \rfloor = 0$. It follows that $\frac{n}{2} + \frac{n}{2^2} + \dots + \frac{n}{2^k} \ge \lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{2^2} \rfloor + \dots + \lfloor \frac{n}{2^k} \rfloor \ge n$.
Then $\frac{n}{2} + \frac{n}{2^2} + \dots + \frac{n}{2^k} \ge n$, so $1 - \frac{1}{2^k} \ge 1$, which is false.

If $a \ge b+2$, then $a! + b! = b! \cdot (1 + (b+1) \cdot (b+2) \cdot \dots \cdot a) = n \cdot 2^a$ and, since $1 + (b+1) \cdot (b+2) \cdot \dots \cdot b$ is odd, we get $2^a \mid b!$. But $b! \mid a!$, so $2^a \mid a!$, which is false.