69th Belarusian Mathematical Olympiad

Answer: not necessarily. Let us show that the function $f(n)=n+n(n-1)+n(n-1)(n-2)+\cdots +n!$ satisfy the conditions of the problem. For any positive integers $m$ and $n$ ($m>n$) the value of $f(m)$ equals to the sum $$m+m(m-1)+\cdots +m(m-1)\dots (m-(n-1))+S(m,n),$$ where $S(m,n)$ is the sum of numbers, divisible by $m-n$. Consider the polynomial $q(x)=x+x(x-1)+\cdots +x(x-1)\dots (x-(n-1))$. Since it has integer coefficients, the difference $q(m)-q(n)$ is divisible by $m-n$. Therefore the difference $f(m)-f(n)=q(m)-q(n)+S(m,n)$ is divisible by $m-n$ as well.

Suppose that $f(n)$ coincides with some polynomial of degree $k$. By the definition of $f$, the inequality $f(n)>n(n-1)\dots (n-k)$ holds for all positive integers $n$. However, in the right-hand side of this inequality there is a polynomial of greater degree $k+1$ and a positive leading coefficient, hence this inequality is not true for all sufficiently large $n$ — a contradiction.