Q:

Prove there exists a unique polynomial

I'm having trouble proving the following lemma:
Let $ p $ be a prime and $ f \in \mathbb{Z}_p [x_1, \dots, x_n] $.  Prove there exists a unique polynomial $ f^* \in \mathbb{Z}_p [x_1, \dots, x_n] $ such that:
a) $ f((x_1, \dots, x_n)) = f^*((x_1, \dots, x_n)) $ for all $ (x_1, \dots, x_n) \in \mathbb{Z}^n $
b) each variable in every non-zero monomial appearing $ f^* $ has exponent
less than $ p $
c) $ \deg(f^*) \leq \deg(f) $. 
I was trying to prove that by induction with respect to the number of variables.  Unfortunately that led me nowhere and I am stuck at the very beginning

A:

Among all polynomials that have poperties  a) and b) (such polynomials exist, for exampl e$f$ itself), let $f^*$ be one that minimizes the sum of exponentes of nonzero monomials.  Then show that $f*$ has property b).  Indeed, if a monomial $ax_1^{e_1}x_2^{e_2}\cdots x_n^{e_n}$ occurs with $a\ne 0$ and some $e_i\ge p$, we can use the Fermat identity $x^p\equiv x\pmod p$ to observe that the polynomial $g$ obtained from $f^*$ by replacing $ax_1^{e_1}x_2^{e_2}\cdots x_n^{e_n}$ with $ax_1^{e_1}x_2^{e_2}\cdots x_i^{e_i-p+1}\cdots x_n^{e_n}$ has properties a) and c) and thus contradicts the minimality of $f^*$. We conclude that no such monomial exists in $f^*$.
As for uniqueness: Count the number of polynomial functions $\mathbb Z_p^n\to \mathbb Z_p$

