Theorem: Lower and Upper Bound of Factorial

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\dfrac {n^n} {e^{n - 1} } \le n! \le \dfrac {n^{n + 1} } {e^{n - 1} }$

Proof:
We have:
{{begin-eqn}}
{{eqn | l = 1 + x
      | o = \le
      | r = e^x
      | c = Exponential of $x$ not less than $1+x$
}}
{{eqn | ll= \leadsto
      | l = 1 + \frac 1 k
      | o = \le
      | r = e^{1 / k}
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {k + 1} k
      | o = \le
      | r = e^{1 / k}
      | c = 
}}
{{end-eqn}}
and similarly:
{{begin-eqn}}
{{eqn | l = 1 + x
      | o = \le
      | r = e^x
      | c = Exponential of $x$ not less than $1+x$
}}
{{eqn | ll= \leadsto
      | l = 1 - \frac 1 {k + 1}
      | o = \le
      | r = e^{-1 / \left({k + 1}\right)}
      | c = 
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \frac k {k + 1}
      | o = \le
      | r = e^{-1 / \left({k + 1}\right)}
      | c = 
}}
{{end-eqn}}
First the {{LHS}} inequality:
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^{n - 1} \dfrac {\left({k + 1}\right)^k} {k^k}
      | r = \frac {2 \times 3^2 \times 4^3 \times \cdots \times n^{n - 1} } {1 \times 2^2 \times 3^3 \times 4^4 \times \cdots \times \left({n - 1}\right)^{n - 1} }
      | c = 
}}
{{eqn | r = \frac {n^{n - 1} } {\left({n - 1}\right)!}
      | c = as most of it cancels
}}
{{eqn | r = \frac {n \times n^{n - 1} } {n \left({n - 1}\right)!}
      | c = 
}}
{{eqn | n = 3
      | r = \frac {n^n} {n!}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^{n - 1} \dfrac {\left({k + 1}\right)^k} {k^k}
      | o = \le
      | r = \prod_{k \mathop = 1}^{n - 1} \left({e^{1 / k} }\right)^k
      | c = from $(1)$
}}
{{eqn | r = \prod_{k \mathop = 1}^{n - 1} e
      | c = 
}}
{{eqn | r = e^{n - 1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {n^n} {n!}
      | o = \le
      | r = e^{n - 1}
      | c = from $(3)$
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {n!}
      | o = \le
      | r = \frac {e^{n - 1} } {n^n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = n!
      | o = \ge
      | r = \frac {n^n} {e^{n - 1} }
      | c = Ordering of Reciprocals
}}
{{end-eqn}}
{{qed|lemma}}
Now the {{RHS}} inequality:
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^{n - 1} \dfrac {k^{k + 1} } {\left({k + 1}\right)^{k + 1} }
      | r = \frac {1^2 \times 2^3 \times 3^4 \times \cdots \times \left({n - 1}\right)^n} {2^2 \times 3^3 \times 4^4 \times \cdots \times \left({n - 1}\right)^{n - 1} \times n^n}
      | c = 
}}
{{eqn | r = \frac {\left({n - 1}\right)!} {n^n}
      | c = as most of it cancels
}}
{{eqn | r = \frac {n \left({n - 1}\right)!} {n \times n^n}
      | c = 
}}
{{eqn | n = 4
      | r = \frac {n!} {n^{n + 1} }
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \prod_{k \mathop = 1}^{n - 1} \dfrac {k^{k + 1} } {\left({k + 1}\right)^{k + 1} }
      | o = \le
      | r = \prod_{k \mathop = 1}^{n - 1} \left({e^{-1 / \left({k + 1}\right)} }\right)^{k + 1}
      | c = from $(2)$
}}
{{eqn | r = \prod_{k \mathop = 1}^{n - 1} e^{-1}
      | c = 
}}
{{eqn | r = \dfrac 1 {e^{n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {n^n} {n!}
      | o = \le
      | r = \frac {n!} {n^{n + 1} }
      | c = from $(4)$
}}
{{eqn | ll= \leadsto
      | l = n!
      | o = \le
      | r = \frac {n^{n + 1} } {e^{n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Hence the result.
{{qed}}