calculus - Show that $prodlimits_{i=1}^inftyleft(1-frac{alpha}{i}right)=0$ if $alpha > 0$

Show that $\prod\limits_{i=1}^\infty\left(1-\frac{\alpha}{i}\right)=0$ if $\alpha > 0$.

Hint: Look at the logarithm of the absolute value of the product.

My attempt

If $\alpha\in\mathbb{N}$ then $\exists i=\alpha$ so one factor will equal $0$ and thus the desired result will be attained. I now consider the case when $\alpha\notin\mathbb{N}$. I don't know how to go about this so I attempt to use the hint.

$$\prod_{i=1}^\infty\left\vert1-\frac{\alpha}{i}\right\vert=\operatorname{exp}\left(\ln\left(\prod_{i=1}^\infty\left\vert1-\frac{\alpha}{i}\right\vert\right)\right) = \operatorname{exp}\left(\sum_{i=1}^\infty\ln\left(\left\vert1-\frac{\alpha}{i}\right\vert\right)\right)$$

I don't know where to go from there. I assume that I am to show that the series diverges to $-\infty$. I have briefly had a look at series but in the book I'm currently studying I haven't gotten to the series part yet so I don't think the author expects me to use a bunch of fancy methods to deduce the divergence of the series.