Let $X_n$ be a sequence of iid random variables with $X_n\in L^1$. Prove
$$\lim_{n\to \infty} \left( \prod^n_{i=1}e^{X_i}\right)^{\frac{1}{n}}=e^{EX_1}\quad \text{almost surely}$$
We have $$ \left( \prod^n_{i=1}e^{X_i}\right)^{\frac{1}{n}}=\left( e^{\sum_{i=1}^nX_i}\right)^{\frac{1}{n}}=\left( e^{\sum_{i=1}^nX_i/n}\right)$$
since $e^x$ is continuous we must have $$\lim_{n\to \infty }\sum_{i=1}^n\frac{X_i}{n}=EX_1$$which looks like the strong law of large numbers, how do I bring the almost surely into this?
If $\lim_{n\to\infty} a_n=a$ then $\lim_{n\to\infty} e^{a_n}=e^a$.
So $$\{\lim_{n\to\infty}\frac1n\sum_{i=1}^nX_i=\mathbb EX_1\}\subseteq\{\lim_{n\to\infty}e^{\frac1n\sum_{i=1}^nX_i}=e^{\mathbb EX_1}\}=\{\lim_{n\to \infty} \left( \prod^n_{i=1}e^{X_i}\right)^{\frac{1}{n}}=e^{\mathbb EX_1}\}$$
If the event on LHS has probability $1$ then so has the event on RHS.