For every $m \in \mathbb{N}$ we can choose that way:
$$m = p((p-1)m + 3) = (p-1)(pm + 3) + 3$$
Therefore
$$2S_n = \sum_{k=1}^{p-1} (k^n + (p-k)^n) = \sum_{k=1}^{p-1} \sum_{m=0}^{n-1} C_n^m (p-k)^m p^{n-m} \equiv$$
$$\begin{aligned} & \equiv \sum_{k=1}^{p-1} (-1)^{n-2} \cdot C_n^{n-2} \cdot (p-k)^{n-2} p^2 + (-1)^{n-1} C_n^{n-1} \cdot (p-k)^{n-1} \cdot p \equiv \\ & \equiv (-1)^{n-2} \cdot C_n^{n-2} S_{n-2} p^2 + (-1)^{n-1} C_n^{n-1} S_{n-1} p \pmod{p^3} \end{aligned}$$
Moreover,
$$S_{n-2} = 1^{n-2} + 2^{n-2} + \dots + (p-1)^{n-2} \equiv S_1 = \frac{p(p-1)}{2} \equiv 0 \pmod{p} \ (*)$$
$$S_{n-1} = 1^{n-1} + 2^{n-1} + \dots + (p-1)^{n-1} \equiv S_2 = \frac{p(p-1)(2p+1)}{6} \equiv 0 \pmod{p} \ (**)$$
$p$ is odd prime number hence $(p, 6) = 1$. Thus $2S_n \equiv 0 \pmod{p^3}$. Because, we chose above $p \mid n$ and $n = (p-1)(pm + 3) + 3$. By Fermat's theorem, we get $k^{n-1} \equiv k^2 \pmod{p}$, $k^{n-2} \equiv k \pmod{p}$ then $S_{n-2} \equiv S_1 \pmod{p}$, $S_{n-1} \equiv S_2 \pmod{p}$. Finally, by (*) and (**) proof is completed.