Are large powers of polynomials linearly independent? Let $P_1,\dots,P_k$ be polynomials over $\mathbf{C}$, no two of them being proportional.
Does there exist an integer $N$ such that $P_1^N,\dots,P_k^N$ are linearly independent?
 A: The answer is yes. In fact, an even stronger claim is true: there exists some $N$ such that for all $n \geq N, \ P_{1}^{n}, \dots, P_{k}^n$ are linearly independent over $\mathbb{C}$.
For this we will use a generalization of the Mason-Stother's theorem which appears on the Wikipedia page (though I have taken the special case of the curve $C = \mathbb{P}^{1} (\mathbb{C})$ and written it in slightly different language.):
Let $q_1, \dots, q_{k}$ be polynomials such that $q_1 + \cdots + q_{k} = 0$ and every proper subset of $q_1, \dots, q_{k}$ is linearly independent. Then,
$$\max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right)$$

Now, we can prove the claim by induction on $k$. For $k = 2$ it is obvious. Now, by induction for all $n$ large enough every proper subset of $P_{1}^{n}, \dots, P_{k}^{n}$ is linearly independent. Suppose for contradiction that there exist constants $\lambda_{1}, \dots, \lambda_{k}$ such that
$$\lambda_1 P_{1}^n + \cdots + \lambda_{k} P_{k}^n = 0$$
Letting $q_i = \lambda_i P_{i}^{n}$, notice that $q_1, \dots, q_k$ satisfy the requirements of the lemma (we have assumed that $\lambda_i \neq 0$), and therefore
$$n \leq \max \left\{ \mathrm{deg} \left( q_1 \right), \dots, \mathrm{deg} \left( q_{k} \right) \right\} \leq \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( q_1 \cdots q_{k} \right) \right) - 1\right) = \frac{(k - 1)(k - 2)}{2} \left( \mathrm{deg} \left( \mathrm{rad} \left( P_1 \cdots P_k \right) \right) - 1 \right)$$
but the right hand side is constant, and so for $n$ large we get a contradiction.
