Show that if $a_1,\ldots,a_n$ are elements of a group then $(a_1\cdots a_n)^{-1} =a_n^{-1} \cdots a_1^{-1}$

Question

Show that if $a_1,\ldots,a_n$ are elements of a group then $(a_1\cdots a_n)^{-1} =a_n^{-1} \cdots a_1^{-1}$

I understand how to show it for just two elements $a$ and $b$, but how do I show it for $n$ elements? Is it basically the same proof?

Answer 1

Use induction on $n$. The base case is $n=2$, and assuming the statement is true for the product of $n-1$ elements, it is true for $n$ elements since $$(a_1a_2\cdots a_n)^{-1}=(a_1(a_2\cdots a_n))^{-1}=(a_2\cdots a_n)^{-1}a_1^{-1}$$