For any group $\mathrm{ord}( a_1\circ a_2\circ\dots\circ a_{n-1}\circ a_n)=\mathrm{ord}(a_2\circ a_3\cdots\circ a_{n-1}\circ a_n\circ a_1)$ I would appreciate some advice as how to start with the following problem: 

Show through induction that in every Group G, for all $a_1,a_2,a_3,..., a_n$ that: $$\mathrm{ord}( a_1 \circ a_2 \circ ... \circ a_{n-1} \circ a_n) = \mathrm{ord}(a_2 \circ a_3 \cdots\circ a_{n-1} \circ a_n \circ a_1) $$

Me and some friends have talked about this. if the group was abelian, or if it was said that the operation is commutative, our job would have been done. however, we are pretty lost in this and basically are stuck on the "how do we proceed" department. 
I appreciate any tips. 
 A: Let's prove something more general: $ord(xy) = ord(yx)$.
Suppose $ord(xy)=n$. Consider $(yx)^{n+1}=y(xy)^nx = yx$. This means that $(yx)^n=1$ and so $ord(yx)\le n=ord(xy)$. By symmetry, $ord(yx)\ge ord(xy)$ and so $ord(xy) = ord(yx)$.
Now apply this to $x=a_1$ and $y=a_2 \circ \cdots \circ a_n$.
Edit:
A simpler proof that $ord(xy) = ord(yx)$ is to note that $xy$ and $yx$ and actually conjugate: $yx = z^{-1} (xy) z$, for $z=x$. And conjugate elements have the same order (the proof being along the same lines as above).
A: I guess this is how I would pretend that the proof was by induction:
First do the $n=2$ case, as lhf has already done.  
(I'm going to use $o(g)$ for the order of $g$, just because I think it looks nicer.)
Now suppose the statement is true for some $n\geq 2$ and let $a_1,a_2,\ldots,a_{n+1}\in G$.  Then $a_n a_{n+1}\in G$, so $o(a_1 a_2 \ldots a_n a_{n+1} ) = o( a_1 a_2 \ldots (a_n a_{n+1}) ) = o( a_2 \ldots (a_n a_{n+1}) a_1) = o( a_2 \ldots a_n a_{n+1} a_1)$.  Hence, by induction, the statement is true for all $n\in \mathbb{N}$.
