what is the minimum number of conjugacy classes for a group of order $n!$? i became curious about how one might measure the extent to which a given finite group departs from perfect commutativity. a rough-and-ready index of the degree to which a group $G$ is commutative may be computed as the proportion of ordered pairs $(g_1,g_2) \in G \times G$ for which $g_1g_2 = g_2 g_1$
calculating this index, let us call it $\kappa$, turned out to be simpler than i had anticipated. described in words, the probability of two randomly sampled elements commuting is equal to the number of conjugacy classes divided by the order of the group. 
let $\Gamma (=\Gamma_G)$ denote the set of conjugacy classes of the finite group $G$. for every element $g$ of $G$ belonging to $\gamma \in \Gamma$ the number of elements of $G$ which commute with $g$ is the order of $G_g$, the centralizer of $g$. by a basic result of group theory:
$$
|G_g| = \frac{|G|}{|\gamma|}
$$
thus the probability that a randomly-chosen pair of elements of $G$ commute (sampling with replacement) is:
$$
\kappa(G) = \frac{\sum_{g \in G} |G_g|}{|G|^2} \\
 = \frac{|\Gamma_G|}{|G|}  \\
$$
in particular for symmetric groups we have:
$$
\kappa(S_n) = \frac{P(n)}{n!}
$$
where $P(n)$ is the number of partitions of $n$
question is $P(n)$ the minimum number of conjugacy classes for a group of order $n!$, and if so, how can this be demonstrated?

No. For example, the Mathieu group $M_{10}$ has order $6! = 720$, the same as the order of $S_6$, but $M_{10}$ has $8$ conjugacy classes while $S_6$ has $11$. 
You can find a paper on the general question (for groups of various orders) here. 
