$a,b$ 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 $(ab)^n=e=(ba)^n$?
 A: 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?
A: It is true that $(ab)^n = e = (ba)^n$ but that doesn't mean that $n$ is the order of $(ba)$. We can only say that the order of $(ba)$ divides n. Since we don't know if $n$ is the smaller power $k$ such that $(ba)^k = e$.
To prove that $n$ is also the order of $ba$ you have to complete the proof. Let $k$ be the order of $(ba)$ then:
$(ba)^k=\overbrace{(ba)\cdots (ba)}^{k \text{ times}}=b\overbrace{(ab)\cdots (ab)}^{(k-1)\text{ times}}a=b(ab)^{k-1}a=e\Rightarrow (ab)^{k-1}=(ab)^{-1}\Rightarrow (ab)^k=e$.
Therefore $k$ divides the order of $(ab)$.
But if $k\vert n$ and $n\vert k$ then $n=k$.
