showing $\pi^r$ is irrational 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!
 A: 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!
A: 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.  
