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

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.

The mistake in your work is the conclusion that if $2^n | j * k$, then either $2^n | j$ or $2^n | k$. For instance, consider that 8 divides $12*6 = 72$ but neither 12 or 6 is divisible by 8.