real analysis - Some alternating series converging to values relating to $pi$.

The following series converge to a value relating to $\pi$:
\begin{align}
\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots&=\frac{\pi}{4},\\
\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\cdots&=\frac{\pi^2}{8},\\
\frac{1}{1^3}-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\cdots&=\frac{\pi^3}{32},\\
\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\frac{1}{7^4}+\cdots&=\frac{\pi^4}{96},\\
\frac{1}{1^5}-\frac{1}{3^5}+\frac{1}{5^5}-\frac{1}{7^5}+\cdots&=\frac{5\pi^5}{1536}.
\end{align}

It seems that if we define $$f(n):=\sum_{i=0}^{\infty}\Big(\frac{(-1)^i}{2i+1}\Big)^n,\quad n\in\mathbb{N}_+,$$ then the values of $f$ are related to $\pi$, and in fact I guess we have $$f(n)=A\pi^n,\quad A\in\mathbb{Q}.$$

This is strongly reminiscent of Basel problem, where we have a famous solution based on the Weierstrass factorization theorem. Trying to imitate that proof, I want to find a real function $g$ with $$Z:=g^{-1}(0)=\Big\{\frac{(-1)^i}{2i+1}:i\in\mathbb{N}\Big\},$$ and $g$ can be factorized as something like $$g(x)=\prod_{a\in Z}\Big(1-\frac{x}{a}\Big),$$ and by comparing the Taylor series of $g$ and applying Vieta's formulas and Newton's identities, we might find the value of $f(1)$ or even more. But these are just wild guesses. I haven't even studied complex analysis, and I'm only imitating the proof for Basel problem. I wonder if this leads to any reasonable solution.

My question is: How do we obtain the value of $f(n)$ and how do we prove these? Don't hesitate to post solutions based on complex analysis or more advanced analysis! Thanks in advance.