Solution:

Since $n$ is perfect, we can write
$$2 n = \sum_{1 \leq d \mid n} d = \sum_{0 \leq b_{i} \leq a_{i}} p_{1}^{b_{1}} p_{2}^{b_{2}} \cdots p_{k}^{b_{k}} = \prod_{i=1}^{k}\left(1 + p_{i} + \cdots + p_{i}^{a_{i}}\right)$$
Now assuming $a_{1}$ is odd, we find that
$$\left(1 + p_{1} + \cdots + p_{1}^{a_{1}}\right) \equiv 1 - 1 + \cdots - 1 \equiv 0 \quad (\bmod\ p_{1} + 1)$$
and therefore $p_{1} + 1 \mid 2 n$.
If $p_{1} > 2$, then $p_{1} + 1 \mid n$, but since $p_{1}$ is the smallest prime divisor of $n$, no prime divisor of $p_{1} + 1$ can divide $n$, leading to a contradiction.
We conclude that $p_{1} = 2 \mid n$. Since $p_{1} + 1 = 3 \mid 2 n$, we also get $3 \mid n$. But now note that since $n > 6$, the integers $1, n / 2, n / 3, n / 6$ are distinct, proper divisors of $n$ which sum to $n + 1 > n$, contradicting the fact that $n$ is perfect. We conclude that $a_{1}$ must be even.

Note: The case where $2 \mid n$ can also be solved as follows. The Euler-Euclid Theorem says that if $n$ is an even perfect number, then there exists a prime $p$ such that $n = 2^{p-1}\left(2^{p} - 1\right)$. Since $n > 6$, then $p > 2$ is odd and so $a_{1} = p - 1$ is even.