<s>[INST] Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\
&= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0
\end{align}
 [/INST] If $m<n$, then for all $j=0...m$ we have $1-\frac jm \le 1-\frac jn$ and so $\frac{m-j}{m} \le \frac{n-j}{n}$ (for $j=0$ we have equality) and therefore:
$$
\begin{align}
\left(1+\frac1m\right)^m & = \sum_{k=0}^m \binom{m}{k}\frac{1}{m^k} = \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{m-j}{m}\\
& \le \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n} = \sum_{k=0}^m \binom{n}{k}\frac{1}{n^k}\\
& < \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} = \left(1+\frac1n\right)^n\\
\end{align}
$$
</s>