We can easily see the numbers of terms different from $0$ cant be equal to $1,$ so it is not smaller than $2$ and we'll prove $2$ is satisfied. Assume $n=p_1^{e_1}p_2^{e_2}...p_k^{e_k}$ and consider the polynomial $Q(x)=x^{e_1+e_2+...+e_k}\left ( x^{\varphi (n)}-1 \right ).$ For any $p_i,$ if $p_i \mid x,$ then $p_i^{e_i} \mid x^{e_1+e_2+...+e_k}$ or if $(x,p_i)=1,$ then $p_i^{e_i} \mid x^{\varphi (n)}-1.$ In short, for any $x,$ $p_i^{e_i} \mid Q(x) , \, \forall i=\overline{1,k}$ or $n \mid Q(x).$ Now, if we have $deg(Q(x))=d \leq n,$ then $P(x)=x^{n-d}.Q(x)$ satisfied. So it suffice to prove $e_1+e_2+...+e_n + \varphi (n) \leq n,$ this is an easy ineq.