Let $X$ be strictly positive r.v. such that $X$ not equal to 1 almost surely and $\mathbb{E}(X)=1$. Let $X_1, X_2,...$ be a sequence of i.i.d. random variables each having the same distribution as $X$. Let $M_0=1$ and $$M_n=\exp\left(\sum^n_{i=0} \ln(X_i)\right).$$
How to show that $M_n\to0$ almost surely?
As Shalop mentioned, you want to assume $X_i$ is not a.s. $1$.
$M_n \to 0$ iff $\ln(M_n) \to -\infty$, so you want to look at $$S_n = \sum_{i=0}^n \ln(X_i)$$ Note that $\ln(X_i) \le X_i - 1$, with equality iff $X_i = 1$, so $\mathbb E[\ln(X_i)] < 0$. Now use the Strong Law of Large Numbers.
EDIT: It is possible that $\mathbb E[\ln(X_i)]$ doesn't exist (i.e. the integral diverges to $-\infty$. In that case, replace $\ln(X_i)$ by a truncated version with a finite (and still negative) expectation.
Now SLLN says almost surely $$ S_n/(n+1) \to \mathbb E[\ln(X_i)] < 0$$ In particular, for some $\epsilon > 0$ we have almost surely $S_n/(n+1) < -\epsilon$ for all sufficiently large $n$. And then almost surely $S_n < -(n+1) \epsilon \to -\infty$.
