TITLE: Is the number of solutions of $\phi(x)=n!$ bounded? If yes, what is its bound?
QUESTION [5 upvotes]: Pillai showed in 1929 that the function $A(n)$ giving the number solutions of the equation $\phi(x)=n$ is unbounded in (S. Pillai, On some functions connected with $\varphi(n)$, Bull. Amer. Math. Soc. 35 (1929), 832–836). I'm interested to know  about bounds on solutions of $\phi(x)=n!$ which is assigned A055506 in OEIS, where it is claimed that if $\phi(x) = n!$, then $x$ must be a product of primes $p$ such that $p - 1$ divides $n!$. It is unclear to me if this allows me to prove that there are finitely many solutions of the equation $\phi(x) = n!$.  Probably an equivalent question is to ask:  is the number of solutions of $\phi(x)=n!$ bounded?  If yes, what is its bound ?
Related question:
https://math.stackexchange.com/q/3747571/156150

REPLY [3 votes]: We have
$$\frac n{\varphi(n)}=\prod_{p\mid n}\bigl(1-p^{-1}\bigr)^{-1}
\le2\prod_{\substack{p\mid n\\p\ge3}}\frac32
=2\prod_{\substack{p\mid n\\p\ge3}}3^{\log_3(3/2)}
\le2\prod_{\substack{p\mid n\\p\ge3}}p^{\log_3(3/2)}
\le2n^{\log_3(3/2)}$$
(where $p$ runs over primes), hence
$$\varphi(n)\le m\implies n\le(2m)^{(1-\log_3(3/2))^{-1}}=(2m)^{\log_23}.$$
Using a larger cut-off $k$ in place of $3$, the same argument gives
$$\varphi(n)\le m\implies n\le(c_km)^{\log_{k-1}k},$$
where
$$c_k=\prod_{p<k}\bigl(1-p^{-1}\bigr)^{-1}.$$
Notice that $\log_{k-1}k\approx1+\frac1{k\log k}$ for large $k$.
I will not go into details, but it is easy to prove by well-known elementary arguments that $c_k=O(\log k)$, hence if we choose $k\approx\log m$, we obtain
$$\varphi(n)\le m\implies n\le c\,m\log\log m$$
for some constant $c$.