Law of large number for a product of uniform iid random variables (stick breaking)

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Let $(X_n)_n$, $n = 1, 2, \dots$, be an iid sequence of random variables uniformly distributed on $(0, 1]$. Set $S_0 = 1$ a.s. and, for $n = 1, 2, \dots$, set $S_n = \prod_{k=1}^n X_k$.

Compare $S_n$ and $\mathbb{E} S_n$ for large values of $n$.


The hint is to use the Strong Law of Large Numbers.

I thought to use a logarithm transformation to get: $$ S_n = e^{\log \prod_{k=1}^n X_k} = e^{\sum_{k=1}^n \log X_k} = e^{-n \frac{1}{n}\sum_{k=1}^n (-\log X_k)}. $$

We know that $-\log X_k \sim \text{Exp}(1)$ and $\mathbb{E}[-\log X_k] = 1$, here we can apply SLLN to get: $$ \frac{1}{n}\sum_{k=1}^n (-\log X_k) \to 1 \text{ as } n \to +\infty \text{ almost surely.} $$

Now, is it correct to say this?

$$ S_n = e^{-n \frac{1}{n}\sum_{k=1}^n (-\log X_k)} \to 0 \text{ as } n \to +\infty \text{ almost surely.} $$

And, moreover, is that a solution? I know that $\mathbb{E} S_n = 2^{-n}$ and so $\mathbb{E} S_n \to 0 \text{ as } n \to +\infty$, so they tend to the same limit, but I don't think this will be enough to get a proper comparison between $S_n$ and $\mathbb{E} S_n$. Any sugguestion?