$f_{i}(P)=P^{(i)}\left(x_{i}\right)$ for arbitrary scalars $x_{0}, \ldots, x_{n}$ is a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$ Fix a field $\mathbb{K}$ and a nonnegative integer $n$. Let $\mathbb{K}_{n}[X]$ be the $\mathbb{K}$-vector space of all polynomials in $X$ over $\mathbb{K}$ that have degree $\leq n$. Consider its dual space $\left(\mathbb{K}_{n}[X]\right)^{*}$.
I know that the linear maps $f_i : \mathbb{K}_{n}[X] \to \mathbb{K}$ defined by $f_{i}(P)=P\left(x_{i}\right)$ form a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$ whenever $x_{0}, \ldots, x_{n}$ are $n+1$ arbitrary distinct scalars.
I also know that the linear maps $f_i : \mathbb{K}_{n}[X] \to \mathbb{K}$ defined by $f_{i}(P)=P^{(i)}\left(0\right)$ form a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$.
I am trying to combine these two facts to prove that

the linear maps $f_i : \mathbb{K}_{n}[X] \to \mathbb{K}$ defined by $f_{i}(P)=P^{(i)}\left(x_{i}\right)$ form a basis for $\left(\mathbb{K}_{n}[X]\right)^{*}$ whenever $x_{0}, \ldots, x_{n}$ are $n+1$ arbitrary distinct scalars.

But I can't seem to find a way. Can anyone give me a hint?

We could apply the same argument that I explain here if we show the following.
Claim: If $f_i(P) = 0$ for all $i$, then $P = 0$.
Proof: Proceed inductively. If $f_0(P) = 0$, then $P$ must have the form
$$
P(x) = (x - x_0)Q(x).
$$
Now, $P'(x)$ is a polynomial in $\Bbb K_{n-1}[X]$ for which the functionals $P \mapsto P^{(i-1)}(x_i)$ for $i = 1,\dots,n$ are zero, so by our inductive hypothesis it must be that $P'(x) = 0$. That is, $\frac {d}{dx}[(x - x_0)Q(x)] = 0$. So, $(x-x_0)Q(x)$ must be a constant polynomial. However, if $Q(x)$ is a polynomial, then $(x-x_0)Q(x)$ can only be a constant polynomial if $Q(x) = 0$. Thus, $Q(x) = 0$, so that we indeed have $P = 0$ as desired.
