The problem was to show that
$$\left(1 + \frac{1}{n+1}\right)^{n+1} \gt \left(1 + \frac{1}{n}\right)^{n}$$
using the AM >= GM inequality.
Let $x_1 = x_2 = \dots = x_{n} = 1 + \frac{1}{n}$ and $x_{n+1} = 1$
Applying the AM >= GM to the $x_i$ we get that
$$\frac{x_1 + x_2 + \dots + x_{n+1}}{n+1} \ge \sqrt[n+1]{x_1x_2\dots x_{n+1}}$$
i.e
$$ \frac{n\left(1+\frac{1}{n}\right) + 1}{n+1} \ge \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^{n}}$$
i.e
$$ \left(1+\frac{1}{n+1}\right) \ge \left(1 + \frac{1}{n}\right)^{\frac{n}{n+1}}$$
Raising both sides to the $(n+1)^{th}$ power gives us the result.
$$\left(1 + \frac{1}{n+1}\right)^{n+1} \gt \left(1 + \frac{1}{n}\right)^{n}$$
using the AM >= GM inequality.
Let $x_1 = x_2 = \dots = x_{n} = 1 + \frac{1}{n}$ and $x_{n+1} = 1$
Applying the AM >= GM to the $x_i$ we get that
$$\frac{x_1 + x_2 + \dots + x_{n+1}}{n+1} \ge \sqrt[n+1]{x_1x_2\dots x_{n+1}}$$
i.e
$$ \frac{n\left(1+\frac{1}{n}\right) + 1}{n+1} \ge \sqrt[n+1]{\left(1 + \frac{1}{n}\right)^{n}}$$
i.e
$$ \left(1+\frac{1}{n+1}\right) \ge \left(1 + \frac{1}{n}\right)^{\frac{n}{n+1}}$$
Raising both sides to the $(n+1)^{th}$ power gives us the result.
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