Multiplying and dividing, we have

$${\left(1+\frac{1}{n}\right)}^{n}={\left(1+\frac{1}{m}\right)}^{m}\prod _{k=m+1}^{n}\frac{{\left(1+\frac{1}{k}\right)}^{k}}{{\left(1+\frac{1}{k1}\right)}^{k1}}$$ 

As was shown in the parent entry, the quotients in the product can
be simplified to give

$${\left(1+\frac{1}{n}\right)}^{n}={\left(1+\frac{1}{m}\right)}^{m}\prod _{k=m+1}^{n}{\left(1\frac{1}{{k}^{2}}\right)}^{k}\left(1+\frac{1}{k1}\right)$$ 

By the inequality for differences of powers,
Hence, we have the following upper bound:
By cross mutliplying, it is easy to see that

$$\frac{{k}^{2}+k}{{k}^{2}+k1}\le \frac{{k}^{2}}{{k}^{2}1}$$ 

and, hence,
Factoring the rational function in the product, terms cancel and we
have

$$\prod _{k=m+1}^{n}\frac{{k}^{2}}{(k+1)(k1)}=\frac{n(m+1)}{(n+1)m}=\frac{n}{n+1}\left(1+\frac{1}{m}\right)$$ 

Combining,