$V = {\rm Span}_K(\pi, \pi^2, \dots)$ is clearly a $K$-vector space. If we let $K = \Bbb{Q}$ temporarily, then every element of $V$ is transcendental as it's a finite linear combination $Q(X), \ X = \pi$ of the $\pi^i$ and thus can't equal an algebraic $a$ or else for the polynomial $P(a) = 0$ over $K$, we'd have $P(Q(X))= 0$ at $\pi$.
But what if we let $K = $ some algebraic exension of $\Bbb{Q}$ or even the field of all algebraic numbers. Can we prove a similar statement?
Edit: The only exception to the first pargraph is $0 \in V$ which is not trans.
Proof. Let $f(X)=a_0+a_1X+\dots+a_{n-1}X^{n-1}+X^n\in F[X]$ be the minimum polynomial of $r$ over $F$. Then $r$ is also algebraic over $\mathbb{Q}[a_0,\dots,a_{n-1}]$, which amounts to say that $\mathbb{Q}[[a_0,\dots,a_{n-1},r]$ is finite dimensional over $\mathbb{Q}[a_0,\dots,a_{n-1}]$. Since $a_i$ is algebraic over $\mathbb{Q}$, we have the chain of finite extensions $$ \mathbb{Q}\subseteq\mathbb{Q}[a_0]\subseteq\mathbb{Q}[a_0,a_1]\subseteq \dots\subseteq\mathbb{Q}[a_0,\dots,a_{n-1}] $$ so also $\mathbb{Q}[a_0,a_1,\dots,a_{n-1},r]$ is finite dimensional over $\mathbb{Q}$. Therefore $r$ is algebraic over $\mathbb{Q}$.