If we work in the extended real numbers, where $\infty$ is a perfectly valid number, then it's clear that for any fixed positive integer $a$, the sequence of partial products, $p_a(N) = \prod_{i=a}^{N} i$, grows without bound. So for each fixed $a$, we have
$$\prod_{i=a}^{\infty}i = \lim_{N \to \infty} p_a(N) = \infty$$
Therefore,
$$\lim_{a \to \infty} \prod_{i=a}^{\infty} i = \lim_{a \to \infty} \infty = \infty$$
In the extended real numbers, this is legitimate convergence to $\infty$.
If we work in the real numbers, we still sometimes make a distinction between an arbitrary divergent sequence and one which "diverges to $\infty$", meaning that given any positive $B$, all but finitely many of the terms of the sequence exceed $B$.
In your example, working in the real numbers, we could say that for a fixed $a$, the sequence $p_a(N)$ diverges to $\infty$ as $N \to \infty$. But $\prod_{i=a}^{\infty}i$ is not a real number, so the expression $\lim_{a \to \infty}\prod_{i=a}^{\infty} i$ does not make any sense in $\mathbb{R}$. You can't talk about a limit of a sequence where the members of the sequence are not elements of the space in which you are working. For this reason, as @rschweib points out, you need to work in the extended reals for the question to make sense.