This is a simple consequence of the Borel-Cantelli lemma. Before starting, I think we are missing the assumption that $X_1, X_2, \ldots$ are nonnegative. First, recall the following fact

$X_n \to 0$ almost surely if and only if $P(X_n > \epsilon \mbox{ infinitely often}) = 0$ holds for every $\epsilon > 0$.

This is a consequence of countable additivity. Next, recall that the Borel-Cantelli Lemma

If $A_1, A_2, \ldots$ are events such that $\sum_{n = 1} ^ \infty P(A_n) < \infty$ then $P(A_n \mbox{ happens infinitely often}) = 0$.

This is true because

$$P(A_n \mbox{ infinitely often}) = P\left(\bigcap_{n = 1} ^ \infty \bigcup_{j = n} ^ \infty A_n\right) \le P\left(\bigcup_{j = n'} ^ \infty A_n\right) \le \sum_{j = n'} ^ \infty P(A_n) \stackrel{n' \to\infty}\longrightarrow 0$$

with the limit being $0$ due to the fact that $\sum_j P(A_n) < \infty$.

So, we have established that what we really need for almost sure convergence is that $P(X_n > 0)$ be summable. Decaying exponentially is much stronger than being summable; we could get away with much less.

This is a simple consequence of the Borel-Cantelli lemma. Before starting, I think we are missing the assumption that $X_1, X_2, \ldots$ are nonnegative. First, recall the following fact

$X_n \to 0$ almost surely if and only if $P(X_n > \epsilon \mbox{ infinitely often}) = 0$ holds for every $\epsilon > 0$.

This is a consequence of countable additivity. Next, recall that the Borel-Cantelli Lemma

If $A_1, A_2, \ldots$ are events such that $\sum_{n = 1} ^ \infty P(A_n) < \infty$ then $P(A_n \mbox{ happens infinitely often}) = 0$.

This is true because

$$P(A_n \mbox{ infinitely often}) = P\left(\bigcap_{n = 1} ^ \infty \bigcup_{j = n} ^ \infty A_j\right) \le P\left(\bigcup_{j = n'} ^ \infty A_j\right) \le \sum_{j = n'} ^ \infty P(A_j) \stackrel{n' \to\infty}\longrightarrow 0$$

with the limit being $0$ due to the fact that $\sum_j P(A_j) < \infty$.

So, we have established that what we really need for almost sure convergence is that $P(X_n > \epsilon)$ be summable for every $\epsilon$. Decaying exponentially is much stronger than being summable; we could get away with much less.