After proving the $n=2$ case, a lecturer I'm watching stated the generalization of Bezout's lemma and left the induction as an exercise. I think I can see how to prove it without induction, but with it I'm having some trouble. The full statement is:
Let $n \geq 2$, and let $a_1, a_2 \ldots, a_n$ be integers, at least one of which are non-zero. Thus there exist integers $x_1, x_2 \ldots, x_n$ such that $$ \gcd(a_1, a_2, \ldots, a_n) = a_1 x_1 + a_2 x_2 + \ldots + a_n x_n. $$
I already have the $n = 2$ case, so I will assume that the $k$ case, holds meaning $$ \gcd(a_1, \ldots, a_k) = \sum\limits_{i=1}^k a_i x_i. $$ Now I need to be able to say something about $\gcd(a_1, \ldots, a_k, a_{k+1})$. I am not sure if I can say $$ \gcd(\gcd(a_1, \ldots, a_k), a_{k+1}) = \gcd(a_1, \ldots, a_k, a_{k+1}). $$ If I could, the result follows from the $n = 2$ case. It seems plausible to me. The quantity on the left side has to divide $\gcd(a_1, \ldots, a_k)$, which divides $a_1, \ldots, a_k$, so the left side divides each of $a_1, \ldots, a_k, a_{k+1}$. The left-hand side therefore has to divide the right-hand side. As for the right-hand side, since it divides $a_1, \ldots, a_k$, it divides $\gcd(a_1, \ldots, a_k)$; it also divides $a_{k+1}$, so it must divide the right-hand side. As two positive quantities divide each other, they must be equal.
How does this look? I can't think of another way to use the induction hypothesis. There's a longer proof I've seen which proceeds similarly to the $n = 2$ case, but the lecturer called the induction "cheating" in some sense, so I assume there was some kind of key insight that wasn't coming to me immediately.
