**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.1 Exercise 1.1.15**

Let $G$ be a group. Prove that $$(a_1 \cdot \ldots \cdot a_n)^{-1} = a_n^{-1} \cdot \ldots \cdot a_1^{-1}$$ for all $n \in \mathbb{Z}^+$ and $a_i \in G$.

Solution: For $n = 1$, note that for all $a_1 \in G$ we have $a_1^{-1} = a_1^{-1}$.

Now for $n \geq 2$ we proceed by induction on $n$. For the base case, note that for all $a_1, a_2 \in G$ we have $$(a_1 \cdot a_2)^{-1} = a_2^{-1} \cdot a_1^{-1}$$ since $$a_1 \cdot a_2 \cdot a_2^{-1} a_1^{-1} = 1.$$ For the inductive step, suppose that for some $n \geq 2$, for all $a_i \in G$ we have $$(a_1 \cdot \ldots \cdot a_n)^{-1} = a_n^{-1} \cdot \ldots \cdot a_1^{-1}.$$ Then given some $a_{n+1} \in G$, we have \begin{align*} (a_1 \cdot \ldots \cdot a_n \cdot a_{n+1})^{-1} = &\ \left( (a_1 \cdot \ldots \cdot a_n) \cdot a_{n+1} \right)^{-1}\\ = &\ a_{n+1}^{-1} \cdot (a_1 \cdot \ldots \cdot a_n)^{-1}\\ = &\ a_{n+1}^{-1} \cdot a_n^{-1} \cdot \ldots \cdot a_1^{-1}, \end{align*} using associativity and the base case where necessary.