Show that \sqrt[3]{\pi} is transcendental

Step 1
Suppose $a=\sqrt[n]{\pi }$ is the root of a polynomial $P\left(x\right)$ with rational coefficients of positive degree.
Writing ${\displaystyle P\left(a\right)=0}$ and using that ${\displaystyle a^n=\pi }$, powers ${\displaystyle a^k}$ with ${\displaystyle k>n}$ can be replaced with $a^k={\pi }^{\frac{k}{n}}·{\sqrt[n]{\pi }}^{k\mathrm{mod}n}$
It follows that there exist rational polynomials in $\pi a_j\mid {,}_{j=0,1,\cdots ,n-1}$ such that:
1) $a_{n-1}{\sqrt[n]{\pi }}^{ n-1}+a_{n-2}{\sqrt[n]{\pi }}^{ n-2}+\cdots +a_1\sqrt[n]{\pi }+a_0=0$
Multiply the above by $\sqrt[n]{\pi }$ successively ${\displaystyle n-1}$ times:
2) $a_{n-2}{\sqrt[n]{\pi }}^{ n-1}+a_{n-3}{\sqrt[n]{\pi }}^{ n-2}+\cdots +a_0\sqrt[n]{\pi }+a_{n-1}\pi =0$
$a_{n-3}{\sqrt[n]{\pi }}^{ n-1}+a_{n-2}{\sqrt[n]{\pi }}^{ n-2}+\cdots +a_{n-1}\pi \sqrt[n]{\pi }+a_{n-2}\pi =0\cdots$
$a_0{\sqrt[n]{\pi }}^{ n-1}+a_{n-1}\pi {\sqrt[n]{\pi }}^{ n-2}+\cdots +a_2\pi \sqrt[n]{\pi }+a_1\pi =0$
Considering the n equations (1) and (2) as a linear system in $1,\sqrt[n]{\pi },{\sqrt[n]{\pi }}^2,\cdots ,{\sqrt[n]{\pi }}^{n-1}$ it is a homogeneous system with a non-trivial solution, so its determinant must be zero. But all coefficients are rational polynomials in ${\displaystyle \pi }$ which implies ${\displaystyle \pi }$ is algebraic. The contradiction means that the original assumption cannot hold true, so ${\displaystyle \sqrt{n}\left\{\pi \right\}}$ is transcendental.
The same line of proof works for the ${\displaystyle n^{th}}$ root of any transcendental number, not just ${\displaystyle \pi }$.
(The above is essentially proving that $P\left(\sqrt[n]{\pi }\right)=0\Rightarrow R\left(\pi \right)=0$ where R is the resultant $R\left(z\right)=\text{res}(P\left(x\right),,x^n-z,,x)$, only without assuming prior knowledge of resultants or other higher algebra results.)