Just a minor note: what about, say $\frac {(a-1)^{n+1}} {(a-3) \cdot n!} \prod_{p=1}^n \frac {1-a^p} {1-a} \to \prod_{p=1}^n N(p,a)$ with $n \to \infty$? From what I have found it gives a much better approximation for $a > 3$, though the exact form of the estimation of the error is not clear.

Just a minor note: from what I have found $\frac {(a-1)^{n+1}} {(a-3) \cdot n!} \prod_{p=1}^n \frac {1-a^p} {1-a}$ gives a much better approximation to $\prod_{p=1}^n N(p,a)$ with $n \to \infty$ for $a > 3$, though the exact form of the error estimation is not clear.