Prove if $a_{1},...,a_{n}$ are elements of a group $G$, then $(a_{1}*a_{2}*...*a{_n})^{-1}=((a_{1})^{-1})*...*((a_{n})^{-1})$

Question

Using induction

Base case: $n=1$ then $(a_{1})^{-1} = (a_{1})^{-1}$ which is correct.

Assume this is true for some k∈Z.

Then for the $k+1$ case

$(a_{1}*a_{2}*...*a_{k} *a_{k+1})^{-1}=(a_{1}*a_{2}*...*a_{k})^{-1} *(a_{k+1})^{-1}$

By the induction hypothesis we then have $(a_{k})^{-1} * ...*(a_{1})^{-1} * (a_{k+1})^{-1}$

And this is where I'm stuck. I need the get the $k+1$ term to the left hand side of the equation. We don't know if the group is Abelian though. So what do I do now?