Let $n > 6$ be a perfect number. Let $p_1^{a_1} \cdot p_2^{a_2} \cdot ... \cdot p_k^{a_k}$ be the prime factorisation of $n$, where we assume that $p_1 < p_2 <...< p_k$ and $a_i > 0$ for all $ i = 1,...,k$. Prove that $a_1$ is even. Remark: An integer $n \ge 2$ is called a perfect number if the sum of its positive divisors, excluding $ n$ itself, is equal to $n$. For example, $6$ is perfect, as its positive divisors are $\{1, 2, 3, 6\}$ and $1+2+3=6$.
Problem
Source: Switzerland - 2022 Swiss Final Round p7
Tags: number theory, Even, prime factorization, Perfect Numbers, perfect number