Inverse of a product in a group can be written as the product of the inverses of each element in reverse order

Question

Let $(G,\circ)$ be a group and let $g_1,...,g_n\in G, n\in\aleph$. Prove that $(g_1\circ ...\circ g_n)^{-1}=g_n^{-1}\circ ...\circ g_1^{-1}$

I tried this by induction but was unsure how to take out the $g_{n+1}$ without actually using what I was trying to prove!

Answer 1

Note that $$(g_1g_2\cdots g_n)(g_n^{-1}g_{n-1}^{-1}\cdots g_1^{-1})=(g_1g_2\cdots g_{n-1}(g_ng_n^{-1})g_{n-1}^{-1}\cdots g_1^{-1})$$ $$=(g_1g_2\cdots g_{n-2}(g_{n-1}g_{n-1}^{-1})g_{n-2}^{-1}\cdots g_1^{-1})=\cdots=(g_1g_1^{-1})=1$$ This can be formalized into an induction proof.