Follow-up question to "Combinatorial Proof of Fermat's Little Theorem" The original problem has been discussed here, which involves a necklace of $p$ beads each of which can be one of $a$ colors and shows that $a^p - a$ must be a multiple of $p$ by classifying the necklaces which are cyclic permutations of each other as indistinguishable and a necklace of length $p$ has $p$ cyclic permutations leading to $\frac{a^p - a}{p}$ distinct necklaces. Please see the original posting for details.
Follow-up question: Since there are $a!$ ways to permute the $a$ colors, is $\frac{a^p - a}{p}$ divisible by $a!$ ? 
Clarification: We should assume $a < p$ if not use $a \bmod p$. This will lead to such necklaces where all $a$ colors used.

Suppose $a=4\,,p=3\,$
$$\frac{4^3-4}{3×4×3×2}$$
$$\frac{64-4}{4×3×3×2}$$
$$\frac{60}{4×3^2×2}$$
$$\frac{5}{6}$$
Hence disproved

The thing is, the necklaces you got may have such combinations where 1 color might have repeated less than $p$ times, $$$$(as you only discarded the ones which had 1 colour repeated $p$ times specifically)$$$$multiplying by $a!$ doesn't count those permutations

