Linear independence of the numbers $\{1,\pi,{\pi}^2\}$

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 $\Bbb C$.

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.