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?
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?
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.
We have used this problem for our Student Olympiad in Algebra at Moscow State University (in Russian, Пятнадцатая олимпиада, задача 8). So, here is a completely elementary solution.
Exercise 1. Show that, if two tuples $a_1,\dots,a_k$ and $b_1,\dots,b_k$ of complex numbers (or complex polynomials) are such that $\sum a_j^n=\sum b_j^n$ for all positive inters $n$, then these tuples are the same up to a permutation. (Hint: the elementary symmetric polynomials can be expressed via the power sums.)
If polynomials $P_j^n$ are linearly dependent for all $n$, then the matrices $$ \pmatrix{ \bigl(P_1(x_1)\bigr)^n&\dots&\bigl(P_1(x_k)\bigr)^n \cr \dots&\dots&\dots \cr \bigl(P_k(x_1)\bigr)^n&\dots&\bigl(P_k(x_k)\bigr)^n \cr } $$ are singular for all positive integer $n$ and all complex $x_j$. Thus, the determinants are zero, i.e., we obtain $$ \sum_{\sigma\in A_k}\left(\prod_{i=1}^kP_i(x_{\sigma(i)})\right)^n= \sum_{\sigma\in S_k\setminus A_k} \left(\prod_{i=1}^kP_i(x_{\sigma(i)})\right)^n $$ (where $S_k$ and $A_k$ are the symmetric and alternating groups).
By Exercise 1, this means (in particular) that $ \prod\limits_{i=1}^kP_i(x_i) = \prod\limits_{i=1}^kP_i(x_{\sigma(i)}) $ for a non-identity (and even odd) permutation $\sigma$. It remains to apply the following simple fact.
Exercise 2. Show that the equality of $s$-variable polynomials $g_1(x_1)g_2(x_2)\dots g_s(x_s)=h_1(x_1)h_2(x_2)\dots h_s(x_s)$ (where $g_i,h_i$ are nonzero complex polynomials) implies that $g_i$ and $h_i$ are proportional for all $i$.
$\require{AMScd} \require{enclose}$EDIT : As noted by Zach Teitler, the argument below only proves that for $m\gg0$, the family $\left\{P_1^{\otimes m}, \dotsc, P_k^{\otimes m} \right\}$ is a free family in: $$ S^{m}\left( \bigoplus_{j=0}^M S^j(V) \right),$$ where $M$ is the maximum degree of the $P_i$. The OP asks if this family remains free after the projection: $$S^{m}\left( \bigoplus_{j=0}^M S^j(V) \right) \longrightarrow \bigoplus_{j=0}^{mM} S^j(V).$$ I won't immediately delete this answer (perhaps later) as someone might have an idea for improving the argument and show that the family remains indeed free after the above-mentionned projection.
I think we can prove the general (multivariate) case using sheaf cohomology (namely Serre's vanishing Theorem) and some basic geometry of the Veronese embedding. In fact, the general statement is the following:
Fact : Let $V$ be a finite dimensional $K$-vector space and let $x_1, \dotsc, x_k$ be distincts points in $\mathbb{P}(V)$. Then for $m$ sufficiently big, the points $v_m(x_1), \dotsc, v_m(x_k)$ generate a $\mathbb{P}^{k-1}$ in $\mathbb{P}\left(S^m(V) \right)$, where: $$ v_m : \mathbb{P}(V) \hookrightarrow \mathbb{P}\left(S^m(V)\right) $$ is the $m$-th Veronese embedding.
Note that $v_m$ is the projectivization of the map $V \longrightarrow S^m(V)$ given by $f \longmapsto f \otimes \dotsb \otimes f$. Hence if $V$ is the vector space of polynomials of degree less than $M$ (where $M$ is the max of the degrees of the $P_k$), the above Fact implies the desired result for the multivariate polynomials $\enclose{horizontalstrike}{P_1, \dotsc, P_k}$.
Proof : We let $J$ the ideal sheaf of the scheme defined by $x_1, \dotsc, x_k$ in $\mathbb{P}(V)$ and $\tilde{J}$ the ideal sheaf of the scheme defined by $v_m(x_1), \dotsc, v_m(x_k)$ in $\mathbb{P}\left(S^k(V) \right)$.
The exact sequence: \begin{CD} 0 @>>> \tilde{J} @>>> \mathcal{O}_{\mathbb{P}\left(S^m(V) \right)} @>>> \mathcal{O}_{v_m(x_1), \ldots, v_m(x_k)} @>>> 0 \end{CD} twisted by $\mathcal{O}_{\mathbb{P}\left(S^m(V)\right)}(1)$ induces an exact sequence in cohomology: \begin{CD} 0 @>>> H^0(\tilde{J}(1)) @>>> H^0(\mathcal{O}_{\mathbb{P}\left(S^m(V) \right)}(1)) @>>> H^0(\mathcal{O}_{v_m(x_1), \ldots, v_m(x_k)}(1)). \end{CD} From this exact sequence, we get that $H^0(\tilde{J}(1))$ is the space of hyperplanes in $\mathbb{P}\left(S^m(V) \right)$ which contain the set of points $v_m(x_1), \ldots, v_m(x_k)$. We will prove that for $m\gg0$ : $$\dim H^0(\tilde{J}(1)) = \dim S^m(V) - k.$$
First consider the commutative diagram: \begin{CD} 0 @>>> \tilde{J} @>>> \mathcal{O}_{\mathbb{P}\left(S^m(V) \right)} @>>> \mathcal{O}_{v_m(x_1), \ldots, v_m(x_k)} @>>> 0 \\ @. @VVV @VVV @| \\ 0 @>>> J @>>> \mathcal{O}_{\mathbb{P}(V)} @>>> \mathcal{O}_{x_1, \ldots, x_k} @>>> 0 \\ \end{CD} The snake's lemma insures that we have an exact sequence: \begin{CD} 0 @>>> K @>>> \tilde{J} @>>> J @>>> 0 \end{CD} where $K$ is the ideal sheaf of $v_m\left(\mathbb{P}(V)\right)$ in $\mathbb{P}\left(S^m(V) \right)$. Twisting this exact sequence by $\mathcal{O}_{\mathbb{P}\left(S^m(V)\right)}(1)$ and taking cohomology, we get an exact sequence: \begin{CD} 0 @>>>H^0(\mathbb{P}(S^m(V)), K(1)) @>>> H^0(\mathbb{P}(S^m(V)),\tilde{J}(1)) @>>> H^0(\mathbb{P}(S^m(V)),J(1)) @>>> H^1(\mathbb{P}(S^m(V)),K(1)). \end{CD} Let us prove that $H^0(K(1)) = H^1(K(1)) = 0$. The exact sequence: \begin{CD} 0 @>>> K @>>> \mathcal{O}_{\mathbb{P}\left(S^m(V) \right)} @>>> \mathcal{O}_{v_m(\mathbb{P}(V))} @>>> 0 \end{CD} twisted by $\mathcal{O}_{\mathbb{P}\left(S^m(V)\right)}(1)$ induces an exact sequence in cohomology: \begin{CD} 0 @>>> H^0(K(1)) @>>> H^0(\mathcal{O}_{\mathbb{P}\left(S^m(V) \right)}(1)) @>>> H^0(\mathcal{O}_{v_m(\mathbb{P}(V))}(1)) @>>> H^1(K(1)) @>>> H^1(\mathcal{O}_{\mathbb{P}\left(S^m(V) \right)}(1)). \end{CD} But we know that $H^1(\mathcal{O}_{\mathbb{P}\left(S^m(V) \right)}(1)) = 0$, that $H^0(\mathcal{O}_{v_m(\mathbb{P}(V))}(1)) = H^0(\mathcal{O}_{\mathbb{P}(V)}(m)) = S^m(V^*)$ and that the induced map: $$ H^0(\mathcal{O}_{\mathbb{P}\left(S^m(V) \right)}(1)) \longrightarrow S^m V^*$$ is the identity. We then find $H^0(K(1)) = H^1(K(1)) = 0$. Hence we get: $$H^0(\mathbb{P}(S^m(V)), \tilde{J}(1)) = H^0(\mathbb{P}(S^m(V)), J(1)).$$ But, by definition of the Veronese embedding $v_m : \mathbb{P}(V) \hookrightarrow \mathbb{P}(S^m(V))$, we have: $$H^0(\mathbb{P}(S^m(V)), J(1)) = H^0(\mathbb{P}(V), J(m)).$$ Furthermore, we have an exact sequence in cohomology: \begin{CD} 0 @>>>H^0(\mathbb{P}(V), J(m)) @>>> H^0(\mathbb{P}(V),\mathcal{O}_{\mathbb{P}(V)}(m)) @>>> H^0(\mathbb{P}(V), \mathcal{O}_{x_1, \ldots, x_k}(m)) @>>> H^1(\mathbb{P}(V),J(m)). \end{CD} We note that $H^0(\mathbb{P}(V), \mathcal{O}_{x_1, \dotsc, x_k}(m)) = K^k$ for all $m$ (since $\{x_1, \dotsc, x_k\}$ is a finite reduced scheme of length $k$), and by Serre's vanishing Theorem, we have $H^1(\mathbb{P}(V),J(m)) = 0$ for all $m\gg0$. Hence, for all $m\gg0$, we have: $$\dim H^0(\mathbb{P}(V), J(m)) = \dim S^m(V) - k,$$ which implies that for all $m\gg0$, we have: $$\dim H^0(\mathbb{P}(S^m(V)), \tilde{J}(1)) = \dim S^m(V)-k.$$ Since $H^0(\mathbb{P}(S^m(V)), \tilde{J}(1))$ is the space of hyperplanes in $\mathbb{P}(S^m(V))$ that contain $\{v_m(x_1), \dotsc, v_m(x_k)\}$, we find that for all $m\gg0$, $v_m(x_1), \dotsc, v_m(x_k)$ generate a $\mathbb{P}^{k-1}$ in $\mathbb{P}(S^m(V))$ and hence form a free family.