Belarusian Mathematical Olympiad

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