Proof. \[\tau(n!) = \prod_p 1+\nu_p(n!) = \prod_p 1+ \frac{n-s_p(n)}{p-1}.\] So, if we knew that these numbers were all distinct, we would be supa-chilling. However, for small \(n\) they actually might not be distinct (e.g., \(n=3,5\)). So we should be careful.

So, we can upper bound these guys using \[\frac{n-s_p(n)}{p-1} \le \frac{n-1}{p-1},\] which gives upper bounds \[1+\frac{n-1}{2-1}, 1+\frac{n-1}{3-1}, 1+\frac{n-1}{5-1}, 1+\frac{n-1}{7-1}\] and lower bound using \[\frac{n-s_p(n)}{p-1} \ge \frac{n}{p-1} - \left\lceil \log_p n \right\rceil,\] which gives lower bounds \[1+ \frac{n}{2-1}-\left\lceil \log_2 n \right\rceil, 1+\frac{n}{3-1}-\left\lceil \log_3 n \right\rceil, 1+\frac{n}{5-1}-\left\lceil \log_5 n \right\rceil.\]

So, can we assert something about distinctness? Let’s compare the bounds on the first two terms; those seem to be a weaker point.

The first factor is at least \(n-\left\lceil \log_2 n \right\rceil + 1\), whereas the second factor is at most \(\frac{n-1}{2}+1.\)

\[\frac{n-1}{2} < n-\left\lceil \log_2 n \right\rceil \iff \left\lceil \log_2 n \right\rceil < \frac{n+1}{2}.\] This is true as long as \(n> 5\).

By similar analysis you can show that for \(n>5\) none of the other terms collide. lit, we win.

Note: just manually verify the \(n\le 5\) stuff.