AM - GM inequality with Bernoulli's inequality

I am stuck on this proof: The AM-GM Inequality is Equivalent to the Bernoulli Inequality.

I get the first part with the Bernoulli Inequality, but how did he get from $$\frac{A_n}{A_{n-1}}\ge \frac{x_n}{A_{n-1}}$$ to $$A_n^n\ge x_{n}A_{n-1}^{n-1}$$
Can someone explain to me what I am missing there?

The initial inequality is not $$\frac{A_n}{A_{n-1}} \ge \frac{x_n}{A_{n-1}}$$ but $$\left(\frac{A_n}{A_{n-1}}\right)^n \ge \frac{x_n}{A_{n-1}}.$$ Multiplying both sides by $$A_{n-1}^n$$, we get $$\left(\frac{A_n}{A_{n-1}}\right)^n A_{n-1}^n \ge \frac{x_n}{A_{n-1}} \cdot A_{n-1}^n \implies A_n^n \ge x_n \cdot A_{n-1}^{n-1}.$$