Why is $\lim_{n \to +\infty }{\sqrt[n]{a_1 a_2 \cdots a_n}} =\lim_{n \to +\infty}{a_n}$

Question

Solving some problems regarding limits and sequence convergence, i stumbled upon a task, and it's solution relies on, and i quote: "We now use a well-known theorem : $$\lim_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} = \lim_{n \to +\infty}{a_n}$$

This isn't really intuitive (at least to me) and I don't know how to prove it. The original task was to find the limit of $$\lim_{n \to +\infty }{\sqrt[n]{\bigg{(}1+\frac{1}{1}\bigg{)} \bigg{(}1+\frac{1}{2}\bigg{)}^2 \ldots \bigg{(}1+\frac{1}{n}\bigg{)}^n}} $$ which of course, using the expression above is just $e$.

Answer 1

Take the logarithm of both sides. Then you want to prove

$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_1^n a_i = \lim_{n\rightarrow \infty} a_n.$$

This is a standard result about Cesàro means.

Answer 2

Take the log of the $n$-root, and applied the Cesaro theorem to it, showing that it will converge to the log of $(a_n)_n$'s limit. Take the exp to finish.