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

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.

Answer 1

We use the following seven binomial-coefficient identities, accompanied by brief proofs. The main proof is farther down.

The first and second identities are $$1 = \binom p0 \qquad 1= \binom pp. \tag{1, 2}$$ Indeed, given a set with $p$ elements, there is precisely $1$ way to select $0$ elements and precisely $1$ way to select all $p$ elements.

The third identity, $$2^p = \sum_{k = 0}^p \binom pk,\tag3$$ is found by replacing $x$ and $y$ each with $1$ in the binomial theorem $$(x + y)^p = \sum_{k = 0}^p \binom pk x^ky^{p - k}.$$

The fourth and fifth identities, $$\binom pk = \binom{p - 1}{k - 1}\frac pk \qquad \binom{p - 1}k = \binom pk - \binom{p - 1}{k - 1},\tag{4, 5}$$ are verified by replacing the binomial coefficients with the their combinatorial definition.

The identities above do not actually require that $p$ be prime, but the next two do. The sixth identity is $$\binom pk \equiv 0 \pmod p \text{ for } 1 \le k \le p - 1.\tag6$$ To prove it, use the combinatorial definition $$\binom pk = \frac{p!}{k!(p - k)!}$$ and note that because $p$ is prime and $1 \le k \le p - 1$, none of the factors in the denominator divides out the factor $p$ in the numerator. Hence, $\binom pk$ is a multiple of $p$ and so congruent to $0$ modulo $p$.

The seventh and final identity, $$\binom{p - 1}{k - 1} \equiv (-1)^{k - 1} \pmod p \text{ for } 1 \le k \le p,\tag7$$ we prove by induction. The claim holds for $k = 1$ because $$\binom{p - 1}{1 - 1} = \binom{p - 1}0 = 1 = (-1)^{1 - 1} \equiv (-1)^{1 - 1} \pmod p.$$ Similarly, the claim holds for $k = p$. Suppose that the claim holds for $k = j$, $1 \le j < p$. Then $$\binom{p - 1}{(j + 1) - 1} = \binom{p - 1}j = \binom pj - \binom{p - 1}{j - 1} \equiv 0 + (-1)(-1)^{j - 1} = (-1)^{(j + 1) - 1} \pmod p$$ where the second equation is the fifth identity, and the congruence uses the sixth identity. Hence, the claim holds for $k = j + 1$ whenever it holds for $k = j$, and so the induction principle guarantees that it holds for every integer $k$, $1 \le k \le p$.

We are now ready for the main proof: \begin{align} \frac{2 - 2^p}p & = \frac1p \left( \binom p0 + \binom pp - \sum_{k = 0}^p \binom pk \right) &&\text{first three identities}\\ & = -\frac1p \sum_{k = 1}^{p - 1} \binom pk \\ & = -\frac1p \sum_{k = 1}^{p - 1} \binom{p - 1}{k - 1} \frac pk &&\text{fourth identity}\\ & = -\sum_{k = 1}^{p - 1} \binom{p - 1}{k - 1} \frac1k \\ & \equiv -\sum_{k = 1}^{p - 1} \frac{(-1)^{k - 1}}k \pmod p &&\text{seventh identity}\\ & \equiv (p - 1)! \sum_{k = 1}^{p - 1} \frac{(-1)^{k - 1}}k \pmod p &&\text{Wilson's theorem}\\ & = \frac{(p - 1)!}1 - \frac{(p - 1)!}2 + \frac{(p - 1)!}3 - \cdots - \frac{(p - 1)!}{p - 1} \end{align}

Remarks: This is Problem 14 of Section 2.7 of I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, 5th ed., Wiley (New York), 1991. See also About the Fermat quotients with base $2$.