Prove that continuous compounding yields a higher return than any other frequency

Let $\displaystyle 0 < r \in \mathbb{R}$ and $\displaystyle n \in \mathbb{N}$. Suppose that we compound at a rate of $\displaystyle r$ at a frequency of $\displaystyle n$ times per year.

Then the return over the course of a year is: $\displaystyle (1 + \frac{r}{n})^n$. If we compound continuously, we obtain a return of $\displaystyle e^r$.

I want to show that continuous compounding yields a higher return than any other frequency of compounding, or that $\displaystyle e^r > (1 + \frac{r}{n})^n$ for all $\displaystyle n \in \mathbb{N}$. I approached this by trying to prove that $\displaystyle (1 + \frac{r}{n})^n < (1 + \frac{r}{n+1})^{n+1}$ for $\displaystyle n \in \mathbb{N}$ and then showing that $\displaystyle e^r$ is the supremum of the series $\displaystyle \{(1 + \frac{r}{n})^n\}$

I am having trouble showing that $\displaystyle (1 + \frac{r}{n})^n < (1 + \frac{r}{n+1})^{n+1}$ - any hints? Thanks!