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What this question is looking for is bounds on $n!$ that are elementary in nature (I seem to have a fetish for these type of proofs).

In general, as the results become more complicated, they also become more accurate, with the ultimate being, of course, Stirling's theorem.

Here are my two candidates:

$$e\left(\dfrac{m}{e}\right)^m < m! < em\left(\dfrac{m}{e}\right)^m $$ and $$\left(\dfrac{m+1}{e}\right)^m < m! < (m+1)\left(\dfrac{m+1}{e}\right)^m $$

Here are their proofs:

We start with $(1+1/n)^n < e < (1+1/n)^{n+1}$, which can be proved in a very elementary manner as shown here: What is the most elementary proof that $\lim_{n \to \infty} (1+1/n)^n$ exists?. Actually, what is proved is that $(1+1/n)^n$ is an increasing sequence, $(1+1/n)^{n+1}$ is a decreasing sequence, and they have a common limit which is called "$e$".

Write this inequality as $\dfrac{(n+1)^n}{n^n} < e < \dfrac{(m+1)^{n+1}}{n^{n+1}}$.

Multiplying, $\prod_{n=1}^m \dfrac{(n+1)^n}{n^n} < e^m < \prod_{n=1}^m \dfrac{(n+1)^{n+1}}{n^{n+1}}$.

For the left side, $\prod_{n=1}^m \dfrac{(n+1)^n}{n^n} =\prod_{n=1}^m \dfrac{(n+1)^n}{n\ n^{n-1}} =\dfrac{(m+1)^m}{m!} =\dfrac{(m+1)^{m+1}}{(m+1)!} $.

For the right side, $\prod_{n=1}^m \dfrac{(n+1)^{n+1}}{n^{n+1}} =\prod_{n=1}^m \dfrac{(n+1)^{n+1}}{n\ n^{n}} =\dfrac{(m+1)^{m+1}}{m!} =\dfrac{(m+1)^{m+2}}{(m+1)!} $.

There are two possible routes from here.

1) $\dfrac{(m+1)^{m+1}}{(m+1)!} < e^m < \dfrac{(m+1)^{m+2}}{(m+1)!} $ or $\dfrac{m^m}{m!} < e^{m-1} < \dfrac{m^{m+1}}{m!} $ or $\dfrac{m^m}{e^{m-1}} < m! < \dfrac{m^{m+1}}{e^{m-1}} $ or $e\left(\dfrac{m}{e}\right)^m < m! < em\left(\dfrac{m}{e}\right)^m $.

2) $\dfrac{(m+1)^{m}}{m!} < e^m < \dfrac{(m+1)^{m+1}}{m!} $ or $\dfrac{(m+1)^m}{e^m} < m! < \dfrac{(m+1)^{m+1}}{e^m} $ or $\left(\dfrac{m+1}{e}\right)^m < m! < (m+1)\left(\dfrac{m+1}{e}\right)^m $.

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