Using mathematical induction to prove Inverse rule for groups elements The question is:

Use mathematical induction to prove that if  $a_1,a_2,\ldots,a_n$ are elements of a group $G$ then $$(a_1a_2\cdots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\cdots a_1^{-1}$$

May attempts:
Starting with a base case where $k=2$ we have $a_1$,$a_2~~$ $(a_1,a_2)^{-1}=e$
we multiply out by $(a_2)^{-1} (a_1)^{-1} a_1,a_2 (a_1,a_2)^{-1}= (a_2)^{-1} (a_1)^{-1} e$
and get 
$e*e*(a_1,a_2)^{-1}=(a_2)^{-1}(a_1)^{-1}$
now i assume true for the $k$-th case and want to show it for the $k+1$ case
${(a_1,a_2,\ldots,a_k)^{-1}} = (a_k)^{-1},\ldots,(a_1)^{-1}$
now i multiply by $(a_{k+1})^{-1}$
$(a_{k+1})^{-1} * {(a_1,a_2,\ldots,a_k)^{-1}} = (a_{k+1})^{-1} * (a_k)^{-1},\ldots,(a_1)^{-1}$
My main question being is can i treat $(a_{k+1})^{-1}$ as one item and ${(a_1,a_2,\ldots,a_k)^{-1}}$ as anther item say $(a_d)^{-1}$ and convert it into  $(a_d,a_{k+1} )^{-1} = (a_{k+1})^{-1} * (a_k)^{-1},\ldots,(a_1)^{-1}$
Does this finish the proof?
 A: Yes, that's precisely the idea. Your inductive step is a bit shakey though. This is how you write it out. You've got the base case, $(ab)^{-1}=b^{-1}a^{-1}$.  For the $k$ case, you can take $a=a_1\cdots a_{k-1}$ and $b=a_k$.  So you've got $$(a_1\cdots a_{k-1}a_k)^{-1}=((a_1\cdots a_{k-1})(a_k))^{-1}=a_k^{-1}(a_1\cdots a_{k-1})^{-1}.$$  By inductive hypothesis $$(a_1\cdots a_{k-1})^{-1}=a_{k-1}^{-1}\cdots a_1^{-1},$$ so $$(a_1\cdots a_{k-1}a_k)^{-1}=a_k^{-1}a_{k-1}^{-1}\cdots a_1^{-1},$$ and you're done.
A: Hint $\ $ Below is the essence of the inductive step, with $\rm\: x' = x^{-1}$
$\rm\begin{eqnarray}\qquad  &&\ \rm (ab\cdots\! &\rm y& &\rm\!\! z)  (z'& &\rm\! y'&\rm\! \cdots b'a')\\
\rm &=&\ \rm  (ab\cdots\! &\rm y&\!\!)&\rm\! z\ \ z'&\!\!(&\rm\! y'&\rm\!\cdots b'a')\quad by\  associativity\\
\rm &=&\ \rm  (ab\cdots\! &\rm y&\!\!)&\ 1&\!\!(&\rm\! y'&\rm\!\cdots b'a')\quad by\, \ zz' = 1\\
\end{eqnarray}$
The prior $ = 1$ by the induction hypothesis, after absorbing the $1$. Ditto in the other order.
