Let $p$ be prime and $n$ be any integer. Suppose $t=(n^{a_n}, \dots, 2^{a_2}, 1^{a_1}) \vdash n$, (i.e. $t$ is a partition of $n$, where we group repeated integers, so, for example, $2^{a_2}$ means that $2$ appears $a_2$ times in the partition).
My question is, can we always find a partition of $n$ (excluding the partition trivial $(1^n)$) such that $p$ does not divide $\frac{n!}{{\prod_{i=1}^n}i^{a_i}a_i!}$?
I have no idea if this is true, but it seems possible.
N.B: I haven't plucked the formula $\frac{n!}{{\prod_{i=1}^n}i^{a_i}a_i!}$ out of thin air: it counts the number of permutations of $[n]$ of cycle shape corresponding to the partition $t$.
Equivalently, can we find a conjugacy class of the symmetric group on $[n]$ whose size is not divisible by $p$?
Let's call the calculated numbers $P(t(n))$, where $t(n)$ is a partition of $n$. It suffices to show that there is no common prime factor of $P(t(n))$ over the non-trivial partitions. But as the OP notes, these numbers count the permutations of $[n]$ of cycle shape corresponding to each $t$, hence their sum (setting the non-trivial partition aside) is $n!-1$. So if they have a common prime factor, it must divide $n!-1$. But one of the $P$'s, namely the one for the partition with $a_n=1$ (and $a_i=0$ otherwise), is equal to $(n-1)!$. Its prime factors all obviously divide $n!$, hence they don't divide $n!-1$, and so we're done.
(Note: The key, for me, to solving this was the OP's comment listing the examples for $n=3$, $4$, and $5$, in particular seeing that the numbers added up to $6$, $24$, and $120$ and then also seeing the numbers $2$, $6$, and $24$ in each list.)