For simplicity let's enumerate the statements.
(1) $(1^{n-1} + 2^{n-1} + \dots + (n-1)^{n-1}) + 1$ is divisible by $n$
(2) $\frac{n}{p}-1$ divisible by $p$ and $\frac{n}{p}-1$ divisible by $p-1$
Let $n = Ap$. Firstly, observe that
$$\sum_{k=1}^{n-1} k^{n-1} \equiv \begin{cases} -A & (\text{mod } p), & \text{if } n-1 \text{ is divisible by } p-1, \\ 0 & (\text{mod } p), & \text{if } n-1 \text{ is not divisible by } p-1 \end{cases}$$
The first case is clear, it follows from Fermat's little theorem. And the second can be shown, for instance, as follows: let $g$ be a primitive root by modulo $p$, then
$$\sum_{k=1}^{p-1} k^{n-1} \equiv \sum_{k=1}^{p-1} (gk)^{n-1} \equiv g^{n-1} \sum_{k=1}^{p-1} k^{n-1} \pmod{p}$$
and $g^{n-1} \not\equiv 1 \pmod{p}$ implies that $\sum_{k=1}^{n-1} k^{n-1} \equiv 0 \pmod{p}$.
Suppose that (1) holds. Then either $n-1$ divisible by $p-1$ and $A-1$ divisible by $p$, or $n-1$ divisible by $p-1$ and $1$ divisible by $p$. Therefore, $n-1$ divisible by $p-1$ and $A-1$ divisible by $p$. Hence, $n-1 = (p-1)A + A-1$ divisible by $p-1 \Rightarrow A-1$ divisible by $p-1$, we obtain (2).
Now suppose that (2) holds. $A-1$ divisible by $p \Rightarrow n$ is not divisible by $p^2$, so $n$ is squarefree. $A-1$ divisible by $p-1 \Rightarrow p(A-1) = n-1-(p-1)$ divisible by $p-1 \Rightarrow n-1$ divisible by $p-1$. Therefore,
$$1 + \sum_{k=1}^{n-1} k^{n-1} \equiv 1 - A \equiv 0 \pmod{p},$$
since $A-1$ divisible by $p$. Because this holds for every prime divisor $p$ of $n$, and $n$ is squarefree, using the Chinese Remainder Theorem, we obtain (1).