Showing that a stopping time is finite for a biased random walk. If we consider $X_i$ iid with $\mathbb{P}(X_i=1) = p$ and $\mathbb{P}(X_i=-1)=1-p$. Where $p \in (1/2,1)$. The random walk is then given by,
$$S_n=\sum_{i=1}^n X_i. $$
We also define the stopping time $\tau = \inf\{k:S_k \in \{-\alpha,\beta\} \}$ with $\alpha,\beta>0$ in the natural numbers.
How to prove that $P(\tau<\infty)=1$? Intuitively it's clear that for sure at some point $S_n$ will hit $\beta$ (because the random walk has a tendincy to move upwards.). I was thinking about showing that $P(S_n=\infty \text{ i.o.})=1$. But I'm not really sure how to make a rigorous argument.
Could someone help me with this?

By the law of large numbers, $\frac1nS_n\to2p-1>0$ a.s. In particular,  $S_n\to\infty$ a.s., and because the steps are $\pm1$ it will a.s. reach any ordinate $\beta\in\mathbb N$ (when starting from $0$).
