Does $N! = 2^m$ hold for any integer values of $N$ and $m$?

Question

For any value of $N$, is it possible that the factorial of $N$ is equal to a power of 2?

Answer 1

If $N \ge 3$ then $3$ will divide $N!$ but $3$ will never divide a power of $2$.

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We can find the prime factorization of $N!$ by noting the following:

If we list the prime numbers in order as $p_1, p_2,p_3,.... $ etc. the there is a specific prime $p_n \le N < p_{n+1}$. So the prime factors of $N!$ are $p_1,...., p_n$. A multiple of prime $p_k$ will appear $\lfloor \frac np_{k} \rfloor$ times so $p_k^{\lfloor \frac np_{k} \rfloor}$ will divide $N!$. Furthermore $p_k^2$ will appear $\lfloor \frac n{p_{k}^2} \rfloor$ times and so on.

So $N! = \prod\limits_{p_k\text{ is prime;}\\p_i \le N}p_k^{(\sum\limits^{i=1\\p_k^i\le N}\lfloor \frac n{p_{k}^i} \rfloor)}$