Even Integer with Abundancy Index greater than 9

Theorem
Let $n \in \Z_{>0}$ have an abundancy index greater than $9$.

Then $n$ has at least $35$ distinct prime factors.

Proof
Since Sigma Function is Multiplicative, it follows easily that abundancy index is multiplicative as well.

We have for any prime $p$ and positive integer $k$:

In fact this is the limit of the abundancy index of a prime power.

The greater the prime, the smaller this value is.

Therefore finding the least number of prime factors a number needs to have its abundancy index exceed $9$ is to find the least $m$ such that:
 * $\displaystyle \prod_{i = 1}^m \frac {p_i} {p_i - 1} > 9$

where $p_i$ is the $i$th prime ordered by size.

The result can be verified by direct computation.