Algebra: find polynomials $M(x)$ and $N(x)$ such that $x^{m}M(x)+(1-x)^{n}N(x)=1$.

Question

Find polynomials $M(x)$ and $N(x)$ such that $$ x^{m}M(x)+(1-x)^{n}N(x)=1. $$ Here are my thoughts about the problem.
If I substitute $0$ in the left side of equation I get $f(0)=N(0)=1$, so I have the last coefficient of $N(x)$ (having defined $f(x)\triangleq x^{m}M(x)+(1-x)^{n}N(x)$).
Now I can take the derivative of $f(x)$ and remember that $f(x)\equiv1$: $$ f'(x)=mx^{m-1}M(x)+x^{m}M'(x)-n(1-x)^{n-1}N(x)+(1-x)^{n}N'(x)\equiv 0. $$ If I substitute $x=0$ as well I get $N'(x)=n$ for the second coefficient, $N''(x)$ seems to be $n(n+1)$ etc.
So after some exploration I can guess that $$ N(x)=1+n+\frac{n(n+1)}{2!}x^{2}+\ldots+\frac{n(n+1)+\ldots+(n+m-2)}{(m-1)!}x^{m-1}. $$ Possibly, I can guess the form of $M(x)$ in the same way, but it seems to me that it is not the right way to get all the solutions of the problem. Can anybody help me with this problem? Thanks in advance!

Answer 1

Hint: Consider the binomial expansion of $1=(x+\overline{1-x})^{m+n-1}$ and note each term in the expansion contains a factor of $x^m$ or $(1-x)^n$.