Once we have found our polynomial, it's easy to convince ourselves that it's the only one matching all points (ignoring polynomials of higher degrees). One way is to try constructing a polynomial by considering the points in a different order - the solution is always the same.
For our proof, we're going to use the method outlined in the articles. This involves a proof about the solutions of polynomials, which shall be discussed first.
Lemma: Any polynomial of degree n has at most n solutions. If it has more than that, it has infinitely many solutions and is thus zero everywhere.
Proof: Take a polynomial p(x) of degree n, with solutions x\-1, x\-2, ..., x\-n. Accordıng to the factor theorem we can write p(x) as the product of n linear factors and a scale factor:
p(x) = a(x - x\-1)(x - x\-2)...(x - x\-n)
Note that any root can be repeated, which leads to squared, cubed etc. terms in the polynomial and does not affect our result.
Now, our assumption is that p(x) has one more solution (say, x\-0) which is different from any of the first n solutions. This means that:
p(x\-0) = a(x\-0 - x\-1)(x\-0 - x\-2)...(x\-0 - x\-n) = 0
We said that x_0 is different from all other solutions. Thus, none of the linear factors can be zero, since there's always a difference between x_0 and any of the x'es. For our expression to be equal to 0, however, at least one factor needs to be zero. That leaves a to be zero.
a = 0
This would mean, however, that the polynomial is always zero, since anything multiplied by our scale factor 0 is also zero.
p(x) = 0(x\-0 - x\-1)(x\-0 - x\-2)...(x\-0 - x\-n) = 0
Now onto the actual proof.
Assume we have two polynomials of degree n (p_1 and p_2) which have more than n points in common. That means:
p_1(x) = p_2(x) for some values of x.
Another way to state this is that the difference between the two polynomials is 0 at those points.
p_1(x) - p_2(x) = 0
(p_1 - p_2)(x) = 0
If we subtract our polynomial, we get a polynomial of degree $\leq n$ (why?). This polynomial must have more than n solutions, however, and thus must be zero everywhere.
p_1 - p_2 = 0
p_1 = p_2
QED.