Multiplying and dividing, we have $$ \left( 1 + {1 \over n} \right)^n = \left( 1 + {1 \over m} \right)^m \prod_{k=m+1}^n { \left( 1 + {1 \over k} \right)^k \over \left( 1 + {1 \over k-1} \right)^{k-1}} $$ As was shown in the parent entry, the quotients in the product can be simplified to give $$ \left( 1 + {1 \over n} \right)^n = \left( 1 + {1 \over m} \right)^m \prod_{k=m+1}^n \left( 1 - {1 \over k^2} \right)^k \left( 1 + {1 \over k-1} \right) $$ By the inequality for differences of powers, $$ \left( 1 - {1 \over k^2} \right)^k < 1 - {k \over k^2 + k - 1} = {(k+1) (k-1) \over k^2 + k - 1} $$ Hence, we have the following upper bound: $$ \left( 1 + {1 \over n} \right)^n < \left( 1 + {1 \over m} \right)^m \prod_{k=m+1}^n {k^2 + k \over k^2 + k - 1} $$ By cross mutliplying, it is easy to see that $$ {k^2 + k \over k^2 + k - 1} \le {k^2 \over k^2 - 1} $$ and, hence, $$ \left( 1 + {1 \over n} \right)^n < \left( 1 + {1 \over m} \right)^m \prod_{k=m+1}^n {k^2 \over k^2 - 1}. $$ Factoring the rational function in the product, terms cancel and we have $$ \prod_{k=m+1}^n {k^2 \over (k+1)(k-1)} = {n (m+1) \over (n+1) m} = {n \over n+1} \left( 1 + {1 \over m} \right) $$ Combining, $$ \left( 1 + {1 \over n} \right)^n < {n \over n+1} \left( 1 + {1 \over m} \right)^{m+1} $$