Consider $C[0,1]$, the space of continuous function on $[0,1]$ with the $\sup$ norm. Let $\Pi_n$ be the finite dimensional subspace consisting of polynomials of degree $n$ or less. Then $L:\Pi_n \to \mathbb{C}$ defined by the right hand side above is a continuous linear functional on this subspace.
Can you extend this functional to a continuous linear functional on $C[0,1]$?
What does the Riesz representation theorem tell you then?
Here is an explanation of why $L$ is continuous:
Suppose $a \in \mathbb{C}^{n+1}$. Then define the polynomial $ p(t) = \sum_{k=0}^n a_k t^k$ (with $a = (a_0,\cdots, a_n)$). Differentiating gives $p'(t) = \sum_{k=0}^{n-1} (k+1)a_{k+1} t^k$, which can be represented by the point $(a_1,\cdots,n a_n, 0)$. So, we can define $D:\mathbb{C}^{n+1} \to \mathbb{C}^{n+1}$ to be the continuous linear operator $D(a) = (a_1,\cdots,n a_n, 0)$. Similarly, if we select $t^*\in [0,1]$, then $p(t^*) = \sum_{k=0}^n a_k (t^*)^k$, which is trivially a linear function of $a$. So let $E_t :\mathbb{C}^{n+1} \to \mathbb{C}$ be the continuous linear functional $E_t(a) = \sum_{k=0}^n a_k t^k$. Then it is clear that the map $\Lambda : \mathbb{C}^{n+1} \to \mathbb{C}$ defined by $\Lambda(a) = \sum_{k=1}^n E_{\frac{k}{n}} D^k (a)$ is also a continuous linear functional.
It remains to show continuity of the map $\pi: \Pi_n \to \mathbb{C}^{n+1}$, where $\pi(p)$ is the unique element of $\mathbb{C}^{n+1}$ such that $p(t) = \sum_{k=0}^n \pi(p)_k t^k$. If $\pi$ is continuous, then it follows that $L = \Lambda \circ \pi$ is continuous. This can be done explicitly using the Vandermonde matrix, or more simply by noting that the polynomials $v_k(t) = t^k$ form a basis for the finite dimensional $\Pi_n$, and define $\pi(v_k) = e_k$. Then $\pi$ can be defined on $\Pi_n$ by extension. Finally, note that $n(p) = \|\pi(p)\|$ defines a norm on $\Pi_n$, and since norms on a finite dimensional space are equivalent, it follows that $n(p) \leq K \|p \|_{\infty}$, where $K>0$ and $\|\cdot \|_{\infty}$ is the $\sup$ norm on $C[0,1]$. Hence the map $\pi$ is continuous.
Note: The extension of the functional $L$ to $C[0,1]$ is only guaranteed to agree with the right hand side above on $\Pi_n$. In particular, for $f \in C[0,1] \setminus \Pi_n$, the function $f$ may not even be differentiable.