Prove that if a Polynomial P(x) of degree n, shares n points with xn, then P(x)=xn.
(a more general proof would be almost the same but for two polynomials, but I think I've proved this if this is proved)
(FTA) Fundamental Theorem of Algebra - Every P(x) of degree n has at most n roots.
The Attempt at a Solution
An idea was to make an analogy between the proof of FTA and this probem. To use for example the factorization: P(x)= ∏(x-ri) Which when proved valid, proves the FTA too. But I don't understand the proofs given online too well.
But also if we accept FTA, we could also possibly analyze P(x)-xn which shall be called D(x).
D(x) has degree (n-1) since xn-xn=0.
Also since P(x) and xn have n equal points, then D(x) has n zeroes.
But FTA says that the D(x) of degree (n-1) has at most n-1 zeroes. But it has n zeroes. Therefore it must equal 0 itself. P(x)-xn=0 ⇔ P(x)=xn
Is this correct? Is there any simpler way to prove or address this or the more general problem?