, 2023

From the binomial formula, we have
$$2^p = C_p^0 + \dots + C_p^p.$$
Since $C_p^0 = C_p^p = 1$, we also have
$$2^p - 2 = C_p^1 + \dots + C_p^{p-1}.$$
For $1 \le k \le p-1$, we will find the value of $C_p^k$ modulo $p^2$. We know that
$$C_p^k = \frac{p!}{k!(p-k)!} = p \cdot \frac{(p-1)!}{k!(p-k)!}.$$
Write $x_k = \frac{(p-1)!}{k!(p-k)!}$. Note that $x_k$ is an integer, because $p x_k = \frac{p!}{k!(p-k)!} = C_p^k$ is clearly an integer, but $p x_k = C_p^k$ is divisible by $p$, as there is no factor of $p$ in the denominator. Since
$$k! = k \cdot (k-1) \cdot \dots \cdot 1 \equiv (-1)^k (p-k) \cdot (p-k+1) \cdot \dots \cdot (p-1) \pmod{p},$$
one has
$$-k x_k \equiv (p-k)x_k \equiv \frac{(p-k)(p-1)!}{(-1)^k (p-1)!(p-k)} \equiv (-1)^k \pmod{p}.$$
Therefore, $(-1)^{k+1} k x_k \equiv 1 \pmod{p}$, which yields $(-1)^{k+1} x_k \equiv a_k \pmod{p}$ and $x_k \equiv (-1)^{k+1} a_k \pmod{p}$. Adding together these congruences, for $k=1, \dots, p-1$, we get $x_1 + x_2 + \dots + x_{p-2} + x_{p-1} \equiv a_1 - a_2 - \dots - a_{p-2} + a_{p-1} \pmod{p}$. Therefore,
$$\begin{aligned} 2^p - 2 &= C_p^1 + \dots + C_p^{p-1} \equiv p(x_1 + x_2 + \dots + x_{p-2} + x_{p-1}) \\ & \equiv p(a_1 - a_2 + \dots + a_{p-2} - a_{p-1}) \pmod{p^2}, \end{aligned}$$
as desired.