Use binomial expansion to show the following inequality

Let $f(x,n):=(1+x/n)^n$, then we need to show $f(x,n+1)\geq f(x,n)$ for $x\geq0,n\geq2$. Observe that the coefficient of $x^k$ in $f(x,n)$ is $$\frac{1}{n^k}\binom{n}{k}=\frac{1}{k!}\cdot\frac{n}{n}\frac{n-1}{n}\frac{n-2}{n}\dotsm\frac{n-k+1}{n}.$$ Correspondingly the coefficient of $x^k$ in $f(x,n+1)$ is$$\frac{1}{(n+1)^k}\binom{n+1}{k}=\frac{1}{k!}\cdot\frac{n+1}{n+1}\frac{n}{n+1}\frac{n-1}{n+1}\dotsm\frac{n-k+2}{n+1}.$$ To solve the problem, all we need to do is to show that the latter is always greater than or equal to the former. To do so, we pair up each of the $k$ factors (excluding $1/k!$) by noting that $$\begin{split}\frac{n}{n}&=\frac{n+1}{n+1}\\ \frac{n-1}{n}&\leq\frac{n}{n+1}\\ \frac{n-2}{n}&\leq\frac{n-1}{n+1}\\ &\vdots\\ \frac{n-k+1}{n}&\leq\frac{n-k+2}{n+1} \end{split}$$ and multiplying all the inequalities together yields the result.

Dividing is a good strategy. If you divide $n\choose k $$\left(\frac{x}{n}\right)^k$ by $n+1\choose k $$\left(\frac{x}{n+1}\right)^k$, you get $\frac{n-k+1}{n+1}\cdot\left(\frac{n+1}{n}\right)^k$. Now write $\frac{n-k+1}{n+1}$ as a telescoping product of $k$ terms of the form $\frac{m}{m+1}$, pair each up with a $\frac{n+1}{n}$ term, and use the fact that $\frac{m}{m+1}\leq\frac{n}{n+1}$ if $m\leq n$.