show that, $\left( \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{2n+1}\right)^k=(k-1)\sum_{n=-\infty}^{\infty}\frac{(-1)^{kn}}{(2n+1)^k}$

Question

I saw this sum on a maths paper, it read "Sophomore's dream". I haven't got a clue what is mean, anyway this is the formula.

(1) $$\left(\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{2n+1}\right)^2=\sum_{n=-\infty}^{\infty}\frac{1}{(2n+1)^2}$$

Using experimental mathematics we was able to arrive at this formula, where k=2 gives (1)

Valid for $k\ge 2$

$$\left( \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{2n+1}\right)^k=(k-1)\sum_{n=-\infty}^{\infty}\frac{(-1)^{kn}}{(2n+1)^k}$$

Question 1,

How do you go about proving this formula? We try the following steps... $$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{2}$$

$$\left( \frac{\pi}{2}\right)^k=(k-1)\sum_{n=-\infty}^{\infty}\frac{(-1)^{nk}}{(2n+1)^k}=\Gamma(k-1)\int_{-1}^1 \frac{(-\ln(x))^{k-1}}{1+x^2}dx$$.

Where $\Gamma(k)$ is the Gamma function is defined as $\Gamma(k)=(k-1)!$

This is where I stop for help. I'm unable to evaluate this integral, can anybody help me, please ?