Linear independence of the numbers $\{1,\pi,{\pi}^2\}$ Does someone know a proof that $\{1,\pi,{\pi}^2\}$ is linearly independent over $\mathbb{Q}$ ?
The proof should not use that $\pi$ is transcendental.
$\{1,e,e^2,e^3\}$  is linearly independent over $\mathbb{Q}$
Any hints would be appreciated.
 A: Edit_1: Indeed, something is wrong. I found the mistake. The main trick used below is not correct:
$$e^{2\pi^2i}=(e^{2\pi i})^\pi=1^\pi=1=e^{2k\pi i}$$
But then $2\pi^2 = 2k\pi$ for some $k\in\Bbb Z$, so that $\pi=k\in \Bbb Z$, which is not possible. I am not exactly sure what is the error. I believe it should be due to $1^\pi =1$, i.e. this is only true in $\Bbb R$ but we are in $\Bbb C$. 

I thought of a possible solution, but I am not sure if I "cheated" somewhere. I guess it is best to post my thoughts.  

Suppose
$$\alpha\pi^2+\beta\pi+\gamma=0$$
for some $\alpha,\beta,\gamma\in \Bbb Q$. Our goal is to prove that $\alpha=\beta=\gamma=0$. Let $D$ be the common denominator of $\alpha,\beta,\gamma$, then multiplying by $D$ we can assume that
$$a\pi^2+b\pi+c=0$$
for some $a,b,c\in \Bbb Z$. Then:
\begin{align*}
2(a\pi+b)\pi i &= -2c i\\
e^{2(a\pi+b)\pi i} &= e^{-2c i}\\
(e^{2\pi i})^{a\pi +b} &= e^{-2c i}\\
1^{a\pi+b} &= e^{-2c i}\\
1 &= e^{-2c i}\\
2k\pi i &= -2c i, \quad k\in \Bbb Z\\
-k\pi &= c
\end{align*}
One possibility is for $k\neq 0$. Then
$$\pi=-\frac{c}{k}\in\Bbb Q$$
However, suppose first that $\pi$ is irrational. Then we must have $k=0$ and hence $c=0$.
Therefore the equation becomes
\begin{align*}
a\pi^2+b\pi &=0\\
a\pi+b &= 0
\end{align*}
since $\pi$ is irrational, once again we must have $a=0$ and $b=0$, which shows that $\{1,\pi,\pi^2\}$ is linearly independent over $\Bbb Q$. Therefore the problem reduces to proving that $\pi$ is irrational. We can refer to the standard proofs for this fact.
