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}}