$\frac{(p-1)!}{1}-\frac{(p-1)!}{2}+\frac{(p-1)!}{3}-\cdots-\frac{(p-1)!}{p-1} \equiv \frac{2-2^p}{p} \pmod{p}$

Question

$p$ is an odd prime. How to prove the following congruence? $$\frac{(p-1)!}{1}-\frac{(p-1)!}{2}+\frac{(p-1)!}{3}-\cdots-\frac{(p-1)!}{p-1} \equiv \frac{2-2^p}{p} \pmod{p}$$ I have created a polynomial that the left side of this congruence is one of its coefficients but this idea not completed to solving the question.