Proof: finite subset closed under product of G is subgroup of G I am reading Herstein on abstract algebra. I have a problem with the following proof:

Now, I understand the first part I think:
As there are more elements $a^k$ with $1 \leq k \leq n+1$ than the size of $H$, there must be $a^i$ and $a^j$ such that $a^i = a^j$. And as $j - i \geq 1 \implies a^{j-i}$ must be some $a^k$. But $a^j = a^i$ so $a^{j-i} = e$. And because as previously shown that $a^{j-i} \in H$ we now know that $e \in H$.
The second part now I do not quite follow; he states that because $j - i - 1 \geq 0$, $a^{j-i-1}$ has to be in $H$... Why? We know $a \in H$ (because $H$ is not empty), and then of course iterating $a$ with the product how many times we want is in $H$ as well, but that is $a^k$ with $k \in \mathbb{N}\setminus{\{0\}}$. So why would $a^0$ be guaranteed to be in $H$. That is what he is saying there.
 A: Try a possibly simpler proof: 
Given $a\in H$, we know that 


*

*$x\mapsto ax$ is a map $H\to H$ because $H$ is closed under multiplication.

*This map is injective because $G$ is a group, more specifically, $ax=ay$ implies $x=y$ because we can multiply with $a^{-1}\in G$ from the left. 

*Because $H$ is a finite set, every injective map $H\to H$ is in fact a bijection.
In particular, there exists $x\in H$ such that $ax=a$. It follows that $x=e$. But then there also exists $y\in H$ such that $ay=e$. It follows that $y=a^{-1}$.

A: Well, $a^{j-i}$ is a power of $a$.
Since $j-i>0$ by assumption, it is an element of $H$, because all powers $a^k\in H$, when $k>0$ (easy proof by induction using closure).
Since we already know $a^{j-i}=e$, we have $e\in H$.
The second part is about showing $a^{-1}\in H$.
Suppose first that $j-i=1$. Then $e=a^{j-i}=a$; in this case, $a^{-1}=e=a\in H$.
Otherwise $j-i>1$; if $k=j-i-1$, then $k>0$ and you can write
$$
e=aa^{j-i-1}=aa^k\in H
$$
Therefore $a^{-1}=a^k\in H$, because $H$ contains all powers of $a$ with positive exponent.

I prefer a different proof, which uses the pigeonhole principle in a different way.
By closure, for every $a\in H$, the map $\rho_a\colon H\to H$, $\rho_a(x)=ax$ is well defined.
By the group properties, $\rho_a$ is injective: if $\rho_a(x)=\rho_a(y)$, then $ax=ay$ and, multiplying by $a^{-1}$ (which exists in $G$), we get $x=y$. Since $H$ is finite, $\rho_a$ is surjective as well. Hence there exists $x\in H$ with $\rho_a(x)=ax=a$; therefore $e=x\in H$. There also exists $y\in H$ with $\rho_a(y)=ay=e$, so $a^{-1}=y\in H$.
Hence we have proved, only using non emptiness, closure and finiteness of $H$ that $e\in H$ and that, for every $a\in H$, $a^{-1}\in H$.
