I want to prove the following statement:
Let $p\gt 3$ be a prime number and let: $$\sum_{j=1}^{p-1}\frac{(-1)^{j}}{j} \binom{p-1}{j} =\frac{a}{b}$$ Which $a,b\in \mathbb Z$ and $\gcd(a,b)=1$. Prove that: $p^2\mid a$.
My idea is the following:
For every $n\in \mathbb N$ and every real $x$ which $|x|<1$, we have :
$$\sum_{j=0}^{n}x^j=\frac{1-x^{n+1}}{1-x}$$ By integration from both side:
$$ \begin{align} \sum_{j=1}^{n+1}\frac{x^j}{j}&=\int_0^x\frac{1-t^{n+1}}{1-t}dt\\ &=-\ln(1-x)-\int_{0}^{x}\frac{t^{n+1}}{1-t}dt\\ &=-\ln(1-x)-\int_{1-x}^{1}\frac{(1-z)^{n+1}}{z}dz\\ &=-\ln(1-x)-\int_{1-x}^{1}\frac{1}{z}(\sum_{j=0}^{n+1}(-1)^{n+1-j}\binom{n+1}{j}z^j)dz\\ &=-\ln(1-x)+\ln(1-x)-\sum_{j=1}^{n+1}(-1)^{n+1-j}\binom{n+1}{j}(\int_{1-x}^{1}z^{j-1}dz)\\ &=\sum_{j=1}^{n+1}\frac{(-1)^{n+1-j}}{j}\binom{n+1}{j}\{(1-x)^{j}-1\}\\ &=(-1)^{n+1}\sum_{j=1}^{n+1}\frac{(-1)^{j}}{j}\binom{n+1}{j}\{(1-x)^{j}-1\} \end{align} $$
Since the equality is valid for $|x| < 1$, and both side are polynomial, so the equality is true for every $z\in \mathbb C$. So if we put $x=1$ and $n=p-2$ then:
$$\sum_{j=1}^{p-1}\frac{(-1)^j}{j} \binom{p-1}{j}=-\sum_{j=1}^{p-1}\frac{1}{j}$$
So, when:
$$\sum_{j=1}^{p-1}\frac{(-1)^j}{j} \binom{p-1}{j}=\frac{a}{b}$$
We have: $$\sum_{j=1}^{p-1}\frac{1}{j}=-\frac{a}{b}$$
But by the Wolstenholme's theorem we know , when:
$$\sum_{j=1}^{p-1}\frac{1}{j}=-\frac{a}{b}$$
Then, $p^2\mid a$.
Is there another solution to this question without using Wolstenholme's theorem ?