About the Fermat quotients with base $2$ On Wikipedia, about Fermat quotients, it says: "Eisenstein discovered that the Fermat quotient with base $2$ could be expressed, $\mod p\ \ $ $p$ odd prime, in terms of the sum  of the reciprocals  of the numbers lying in the first half of the range ${1, p − 1 }$":
$$ -2 \cdot \frac{2^{ p − 1 }-1}{p} \equiv 1+\frac{1}{2}+\cdot\cdot\cdot\cdot+\frac{1}{(p - 1)/2} \mod p  $$
How is this proven?
 A: We use the following seven binomial-coefficient identities, accompanied by brief proofs.  The main proof is farther down.
The first identity, $$2^p = \sum_{k = 0}^p \binom pk, \tag1$$ 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 second and third identities are $$1 = \binom p0 \qquad 1= \binom pp. \tag{2, 3}$$  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 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.  We first show that the expression involving the Fermat quotient is congruent modulo $p$ to the $p - 1$-st "alternating" harmonic number.  Next, we use straightforward algebra and the congruence identities $a \equiv -(p - a)$ and $p - a \equiv -a$ (mod $p$) to show that that alternating harmonic number, in turn, is congruent modulo $p$ to the $(p - 1)/2$-th harmonic number, the desired result.
\begin{align}
-2 \cdot \frac{2^{p − 1} - 1}p & = - \frac 1p(2^p - 2)\\
& = -\frac 1p \left( \sum_{k = 0}^p \binom pk - \binom p0 - \binom pp \right) \qquad \text{first three identities}\\
& = -\frac 1p \sum_{k = 1}^{p - 1} \binom pk\\
& = -\frac 1p \sum_{k = 1}^{p - 1} \binom{p - 1}{k - 1} \frac pk \qquad \qquad \qquad \quad \; \; \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 \qquad \qquad \; \; \text{seventh identity} \\
& = -\frac11 + \frac12 - \frac13 + \frac14 - \cdots - \frac1{p - 4} + \frac1{p - 3} - \frac1{p - 2} + \frac1{p - 1} \\
& = \left( -\frac11 - \frac13 - \cdots - \frac1{p - 4} - \frac1{p - 2} \right) + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& \equiv \left( -\frac1{-(p - 1)} - \frac1{-(p - 3)} - \cdots - \frac1{-4} - \frac1{-2} \right) \\
& \quad + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \pmod p \\
& = \left( \frac1{p - 1} + \frac1{p - 3} + \cdots + \frac14 + \frac12 \right) + \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& = 2 \left( \frac12 + \frac14 + \cdots + \frac1{p - 3} + \frac1{p - 1} \right) \\
& = 1 + \frac12 + \cdots + \frac1{(p - 3)/2} + \frac1{(p - 1)/2}
\end{align}
