If $(G, \circ)$ is a finite group with identity $e$, prove that there exists a positive integer $m$ such that $a^m=e$ holds for all $a\in G$.
Approach:
Edit
Since $G$ is finite, then G has finite number of elements. Assume $G=\{a_1, a_2, a_3, \dots, a_N\}$ for some positive integer $N$. For some $a_i\in G$, consider the sequence $a_i^1, a_i^2, a_i^3, \dots$. All these elements are in $G$ (closed under binary operation). since G is finite, then there must be repetitions in the above sequence, i.e. $a^j_i=a^k_i$ for some positive integers $j$ and $k$ with $j > k$ (without any loss of generality). i.e $a_i^{j-k}=e$ i.e. $a_i^{n_i}=e$.
Here the problem is proved or not? Please suggest me.