Find a polynomial with evaluation equal to # of partitions of n into at most k parts

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Fix a positive integer $k$. For positive integer $n$ let $p(n;\le k)$ denote the number of partitions of $n$ into at most $k$ parts, and let $p(0;\le k)=1$.
(1) Show that there is a polynomial $P(x)$ s.t. $p(n;\le k)=P(n)$ for all nonnegative integers $n$.
(2) And show that $(-1)^{k-1}P(-n)$ equals the number of partitions of n into exactly $k$ distinct parts.
I'm very confused that if the required polynomial has degree $d$, then it is determined uniquely by $d+1$ values, say $P(0),P(1),\cdots,P(d)$. But it seems that the polynomial works for infinite many $n$. I tried to use Lagrange interpolation, but it didn't work well. So how to figure out that? And I think if I know the answer to problem (1), I can work out problem (2). Thanks!


New idea:

I'm trying to use Proposition 4.3 on page 79 of Combinatorial Reciprocity Theorems (a book draft) by Matthias Beck and Raman Sanyal to prove it.

The proposition says "A sequence $f(n)$ is given by a polynomial of degree $\leqslant d$ iff $(\Delta^m f)(0)=0$ for $m>d$."

Where we can set $(\Delta f)(n)=f(n+1)-f(n)$. Then $f(n)=p(n;\le k)$ in our problem. But I'm chocked to find out $(\Delta^m f)(0)$ now.