Hint

Let $P(x)=\sum_{k=0}^n a_k x^k$. Note that if $a_0=0$, the condition trivially holds. Now let $a_0\neq 0$. Fix a prime $p$, such that $p>|a_0|$ and $p>\sum_{k=0}^n |a_k|$. Observe that, for $m=p-1$ and $n=p$, we have $p\mid P(a^{p-1})$. If $p\mid a$, then we earn a contradiction since $p>|a_0|$. Thus, $p\nmid a$, hence, $a\not\equiv 0\pmod{p}$, and by Fermat, $a^{p-1}\equiv 1\pmod{p}$. This yields, $P(a^{p-1})\equiv P(1)\pmod{p}$. Since $p>\sum_{k=0}^n |a_k|\geqslant |P(1)|$ (by triangle inequality), it follows that $P(1)=0$.