Prove that a uniformly convergent convergent sequence of $N^\text{th}$ degree polynomials must converge to some $N^\text{th}$ degree polynomial So here's the question I'm trying to answer:

Suppose $p_n(x) = \sum_{k=1}^N a_k^{(n)} x^k$ is a sequence of polynomials such that $p_n \to f$ uniformly over $[0,1]$ for some function $f:[0,1] \to \mathbb{R}$.  Prove that $f$ must itself be an $N^\text{th}$ degree polynomial.

I've already shown that if each $a_k^{(n)} \to a_k$, then $p_n(x) \to p(x) = \sum_{k=1}^N a_k x^k$  uniformly (earlier part of the problem).  I'm thinking that there's some way to show that if $p_n \to f$, then $a_k^{(n)}$ converges for each $k$.  This certainly works for $k = 0$, since we can guarantee that the sequence $a_k^{(n)} = p_n(0)$ is Cauchy.  I've gotten stuck in trying to extend this to other coefficients; I'm thinking there's some trick involving subtracting the $a_0^{(n)}$ off and dividing by $x$, maybe some fancy induction along those lines.
Other potentially helpful thoughts: we can guarantee that $f$ is continuous since it is the uniform limit of continuous functions. Remember also that we have a compact domain, so that all of these functions are bounded and achieve their max/min.
Any comments, hints, or answers are very much appreciated. 
 A: A proof that for my criterion is elegant is using facts of functional analysis:
Let $P_N$ the set of all polynomials of degrre $N$, this is a vector space and isometrically isomorphic to $\mathbb{R}^{N+1}$ via
$$\begin{array}{rcl}\Phi: P_{N} &\rightarrow & \mathbb{R}^{N+1}\\ a_0+a_1 x +\cdots+a_N x^N & \mapsto & (a_0,\ldots,a_N).\end{array}$$
Given any norm $\left\|\:\:\right\|$  in $\mathbb{R}^{N+1}$ we define the norm in $P_{N}$ for any $p(x)=a_0+a_1 x+\cdots+a_N x^N$ as
$$\left\|p\right\|:=\left\|(a_0,\ldots,a_N)\right\|.$$
Since $\mathbb{R}^{N+1}$ is complete, then $P_N$ is complete with this norm, therefore $(P_N,\left\|\:\:\right\|)$ is a Banach space, but  we know that all norms in a finite-dimensional space are equivalent, therefore, $P_N$ is Banach space with any norm, in particular, with the norm
$$\left\|p \right\|_{\sup}:=\sup_{x\in[0,1]}\left|a_0+a_1 x+\cdots+a_N x^N \right|.$$
From the real analysis courses we know that if $\left\{p_n\right\}$ converges uniformly then it does so in the norm $\left\|\:\right\|_{\sup}$, therefore, $\left\{p_n\right\}$ is a Cauchy sequence in the Banach space $(P_{N}, \left\|\:\right\|_{\sup})$, which allows us to conclude that $f\in P_N$.
A: You may start by sampling the points $0/n, 1/n, ..., (n-1)/n$, and make a vector $$\left( \begin{array}{c} p_m(0) - f(0)\\p_m\left(1/N\right) - f(1/N)\\ \vdots \\ p_m\left(N/N\right) - f( N/N) \end{array} \right) = \left(\begin{array}{cccc} 1& 0& \cdots& 0\\ 1& (1/N)& \cdots & (1/N)^N\\ \vdots& \vdots& \ddots& \vdots\\ 1& N/N& \cdots& ( N/N )^N \end{array}\right)\left( \begin{array}{c}a_0^{(m)}\\a_1^{(m)}\\ \vdots\\ a_N^{(m)} \end{array}\right) - \left( \begin{array}{c}f(0)\\f(1/N)\\ \vdots\\ f(N/N) \end{array}\right)$$
Let us denote this matrix by $V$, the polynomial vector $\vec{p}_m$, the vector with $f$ as $\vec{f}$, and the coefficient vector $\vec{a}_m$. The matrix is called a Vandermonde matrix, and it is invertible. Thus $$\vec{a}_m = V^{-1} (\vec{p}_m - \vec{f}) + V^{-1}\vec{f}$$ and we know that $\vec{p}_m - \vec{f} \to \vec{0}$ as $m\to \infty$. This means that $\vec{a}_m \to V^{-1} \vec{f}$, since every linear operator on a finite dimensional vector space is continuous.
Now we know that the coefficients converge to some vector $\vec{a} = (a_0, a_1, ..., a_N)^T$, and we can demonstrate that $p_m(x) \to p(x) = a_N x^N + \cdots + a_0$.
