Let $x_1, x_2, \ldots, x_n \in G$ for some group $G$. We wish to prove that $$(x_1 x_2 \cdots x_n)^{-1} = x_n^{-1} x_{n-1}^{-1} \cdots x_2^{-1} x_1^{-1}.$$ I'm not sure if the correct way to proceed is by showing the multiplication out, which doesn't seem to me to be required for the inductive step. Here is what I have so far.
Proof. Let $x_1, x_2, \ldots, x_n \in G$ for some group $G$. We proceed by induction on $n$. When $n = 1$, we have $$x_1^{-1} = x_1^{-1}.$$ Less trivially, when $n = 2$, we have $$\begin{align} (x_1 x_2)(x_2^{-1} x_1^{-1}) &= x_1 (x_2 x_2^{-1})x_1^{-1} \\ &= x_1 e x_1^{-1} \\ &= (x_1 e)x_1^{-1} \\ &= x_1 x_1^{-1} \\ &= e, \end{align}$$ and $$\begin{align} (x_2^{-1} x_1^{-1})(x_1 x_2) &= x_2^{-1} (x_1^{-1} x_1)x_2\\ & = x_2^{-1} e x_2 \\ &= x_2^{-1} (ex_2) \\ &= x_2^{-1} x_2 \\ &= e, \end{align}$$ so $(x_1 x_2)^{-1} = x_2^{-1} x_1^{-1}$. Supposing inductively that the result holds when $n = k$, $$ (x_1 x_2 \cdots x_k)^{-1} = x_k^{-1} x_{k-1}^{-1} \cdots x_2^{-1} x_1^{-1},$$ we prove the result when $n = k + 1$: \begin{align*} (x_1 x_2 \cdots x_k x_{k+1})^{-1} & = ((x_1 x_2 \cdots x_k)x_{k+1})^{-1} = x_{k+1}^{-1} (x_1 x_2 \cdots x_k)^{-1} \\ & = x_{k+1}^{-1} (x_k^{-1} x_{k-1}^{-1} \cdots x_2^{-1} x_1^{-1}) \\ & = x_{k+1}^{-1} x_k^{-1} x_{k-1}^{-1} \cdots x_2^{-1} x_1^{-1}. \end{align*}
How does this look?
Your proof is fine.
Nitpicking: your use of associativity misses a few steps; also, it suffices in a group to check whether a candidate inverse of an element is a one-sided inverse for it to be an inverse.