$a,b$ are elements of the group $G$. Why $|ba|\leq |ab|$ in the following scenario? A scenario:
The order, $n$, of $ab$ (and hence, by definition, the order of the cyclic subgroup $\langle ab\rangle$) is finite (thus, the order of $ba$ is finite). Then $(ab)^n=e$. So 
$(ab)^n=\overbrace{(ab)\cdots (ab)}^{n \text{ times}}=a\overbrace{(ba)\cdots (ba)}^{(n-1)\text{ times}}b=a(ba)^{n-1}b=e\Rightarrow (ba)^{n-1}=(ba)^{-1}\Rightarrow (ba)^n=e.$
Therefore, $|ba|\stackrel{*}{\leq} |ab|$. 
Why $*$? Can't we say directly that $(ab)^n=e=(ba)^n$?

I think it's because you have shown that if the order of $ab$ is $n$, then $(ba)^n=e$. This is not the same as concluding that the order of $ba$ is $n$. For that to be the case, it must also be shown that no smaller power will also do the job. I think you can only conclude that the order of $ba$ divides $n$.
After all, it is also true that $(ab)^{7n}=e$, but that doesn't mean that the order of $ab$ is $7n$, right?
