Vieta's might be practical

Since $a_1$ , $a_2$ , ..., $a_5$ are the roots of the summation, which means: $\begin{aligned} \sum_{m=1}^5 mx^m & = 5 \prod_{m=1}^5 (x-a_m) \end{aligned}$

Now, we have:

$\begin{aligned} \prod_{m=1}^5 (4a_m + 1) & = 4^5 \prod_{m=1}^5 \left(a_m + \frac{1}{4} \right) \\ & = - 4^5 \prod_{m=1}^5 \left(-\frac{1}{4} - a_m \right) \\ & = - \frac{4^5}{5} \sum_{m=1}^5 m\left(-\frac{1}{4} \right)^m \\ & = \frac{1}{5} \sum_{m=1}^5 (-4)^{5-m} m \\ & = \frac{165}{5} = \boxed{33} \end{aligned}$