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

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

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

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