Almost sure Convergence of $\lim(\prod_{i=1}^{n} X_i)^{1/n}$

Question

Let $(X_n)_{n \geq 1} $ be a sequence of iid random variables uniformly distributed on the interval [1, 2]. I want to show that it exists a real number $c$ s.t. $ \lim_{n \to \infty} \left(\prod_{i=1}^{n} X_i \right)^{1/n} = c $ almost surely.

To do so, I got the hint that I sould show that the following holds:

For a deterministic sequence $(x_n)_{n\geq 1}$ it holds that $$\lim_{n \to \infty} \left( \prod_{i=1}^{n} x_i \right)^{1/n} = c \iff \lim_{n \to \infty} 1/n \sum_{i=1}^{n} \log(x_i) = \log(c) $$

Questions:

1. Why is this useful/helpful to show the hint?
2. How can the hint be shown?
3. What has to be concluded by the hint in order to answer the "real" question?

Edit: I tried to show this as following:

Let $\log(c) = \frac{\sum \log(x_i)}{n}$. Since the random variables $\log(x_1), ..., \log(x_n)$ are iid, we can apply the SLLN, thus: $\log(c) = \frac{\sum \log(x_i)}{n}$ converges a.s. to $\mathbb E(\log(x_1))$, which equals to: $\frac{1}{2} (\log(1) + \log(2))$.

Is this correct until now? How to conclude?

Answer 1

1 : The hint gives you an equivalent condition to $(\prod X_i(\omega))^{1/n}$ converges to $c$ and the sequence you need to check looks like the SLLN.

2 : It goes without saying that the $(x_n)$'s are taken $>0$. I suggest you take the logarithm of the product.

3 : Once you have this hint your question is barely solved, $$ \mathbb{P} \left( (\prod X_i)^{1/n} \rightarrow c \right ) = \mathbb{P} \left( 1/n \sum \ln(X_i) \rightarrow \ln(c) \right ) $$ what looks again like the SLLN.