Because of a mild allergy to subscripts, I will solve a smaller problem. Let $a,b,c,d$ be distinct numbers, and let $p,q,r,s$ be any numbers. Find a polynomial $P(x)$ of degree $\le 3$ such that $P(a)=p$, $P(b)=q$, $P(c)=r$, and $P(d)=s$.
We use the polynomial
where the numbers $A$, $B$, $C$, and $D$ will be chosen soon.
Note that when we substitute $a$ for $x$ in our expression for $P(x)$, the last three terms die. So
We want $P(a)=p$. That is easy to arrange by choosing
In essentially the same way, we can determine $B$, $C$, and $D$. It is no harder to deal with $10$ points, or $100$.
The uniqueness part is standard field theory.
Remark: If we are in a subscript mood, we can write out a solution of the problem for general $n$. Let
Despite appearances, $Q_i(x)$ is a polynomial, indeed one of degree $n-1$. It is the product of all terms $x-x_j$ except $x-x_i$.
where the $A_i$ are constants that will be chosen soon.
If we let
then $P(x)$ has the desired properties. Nice compact proof. But it may be hard to see what is really going on without going through a few concrete examples, or general examples but for a small $n$, like the $n=4$ we dealt with above.