$ \log(n!) = \log(1) + \ldots + \log(n) $.

This sum is $ \int_{1}^{2} \log(2) + \ldots + \int_{n-1}^{n} \log(n) $ $ \geq \int_{1}^{2} \log(t)dt + \ldots + \int_{n-1}^{n} \log(t)dt $ $ = \int_{1}^{n} \log(t) dt = n \log(n) - n +1 $.

Also $ \log(2) + \ldots + \log(n-1) $ $ = \int_{2}^{3} \log(2) + \ldots + \int_{n-1}^{n} \log(n-1) $ $ \leq \int_{2}^{3} \log(t)dt + \ldots + \int_{n-1}^{n} \log(t)dt $ $ \leq \int_{1}^{n} \log(t)dt = n \log(n) - n + 1 $.
So adding $ \log(n) $,
$ \log(1) + \ldots + \log(n) \leq \log(n) + (n \log(n) - n + 1). $

Finally,
$ n \log(n) - n + 1 $ $ \leq \log(1) + \ldots + \log(n)$ $ \leq \log(n) + (n \log(n) - n + 1). $
Taking exponential,
$ n^n e^{-n} e \leq n! \leq n^{n+1} e^{-n} e $.

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Edit [Ok its pointless to avoid pictures]