showing $\pi^r$ is irrational

Question

I want to show $\pi^r$ is irrational, where $r\in \mathbb{Q}-{0}$

Recently, I learn some proof of $e^{r}$ where $r\in \mathbb{Q}-{0}$ is irrational.

It was done by setting \begin{align} f(x) = \frac{x^n(1-x)^n}{n!} \end{align} with some manipulation of calculus.

Employing similar approach can we do this for $\pi^r$?

If you know some references regarding this please make some comments!

Of course any other approach is welcomed!

Answer 1

The OP is presumably asking if there is a proof that doesn't boil down to proving the transcendentality of $\pi$. A useful reference is "Irrationality of $\pi$ and $e$" by Keith Conrad, which includes proofs that $\pi$ is irrational (Theorem 2.1) and that $e^r$ is irrational for any nonzero rational number $r$ (Theorem 5.1), along with a brief discussion of the difference between the two proofs:

Although the proofs of Theorems 2.1 and 5.1 are similar in the sense that both used estimates on integrals, the proof of Theorem 2.1 did not show $\pi$ is irrational by exhibiting a sequence of good rational approximations to $\pi$. The proof of Theorem 2.1 was an “integer between $0$ and $1$” proof by contradiction. No good rational approximations to $\pi$ were produced in that proof. It is simply harder to get our grips on $\pi$ than it is on powers of $e$.

Remark: I found the Conrad paper through a comment at another MSE question here.

Answer 2

If $r=\frac{p}{q}$, was rational, where $p,q$ integers, i.e., $$ \pi^{p/q}=\frac{m}{n}, \quad\text{for some $m,n\in\mathbb N$,} $$ then $$ \pi^p=\frac{m^q}{n^q}, $$ and hence $\pi$ would be satisfy the equation $$ n^qx^p-m^q=0, $$ and hence $\pi$ would be an algebraic number, but $\pi$ is not algebraic, it is a transcendental number!