Let $ P(X_{1}, \dots , X_{n}) \in \mathbb{F}_{q}[X_{1}, \dots , X_{n}] $ such that $ P(x)=0 $ for all $ x \in \mathbb{F}_{q}^{n} $. Prove that there exist polynomials $ P_{1} , \dots , P_{n} \in \mathbb{F}_{q}[X_{1}, \dots , X_{n}] $ such that $$ P(X_{1}, \dots , X_{n})=\sum_{i=1}^{n}P_{i}(X_{1}, \dots , X_{n})(X_{i}^{q}-X_{i}) $$
For $ n=1 $, this follows easily since then we have that $ P(x)=0 \, \forall x \in \mathbb{F}_{q} $ so $ P(X) $ will be divisible by $ \prod_{x \in \mathbb{F}_{q}}(X-x)=X^{q}-X $ and we are done. I imagine for the general case, we should maybe use induction, but I don't know how.
I would appreciate any help. Thank you!
Lets think about the 2-variable case. Let $F(X,Y)$ vanish everywhere on $k^2$ where $k=\Bbb F_q$. If $F(X,Y)$ has a monomial $X^rY^s$ with $r\ge q$ write it as $X^rY^s=X^{r-q+1}Y^s+(X^q-X)X^{r-q}Y^s$. If $s\ge q$ write it as $X^rY^s=X^rY^{s-q+1}+(Y^q-Y)X^rY^{s-q}$. We can then write $F(X,Y)=G(X,Y)+(X^q-X)H_1(X,Y)+(Y^q-Y)H_2(X,Y)$ where also $G(X,Y)=\sum_{i,j=0}^{q-1}a_{i,j}X^iY^j$ vanishes on all of $k^2$. If some $a_{i_0,j_0}\ne0$ there is $x\in k$ with $\sum_{i=0}^{q-1}a_{i,j_0}x^i\ne0$. Then $G(x,Y)=\sum_{j=1}^{q-1}b_j Y^j$ with $b_{j_0}\ne0$. Then there is $y\in k$ such that $G(x,y)=\sum_{j=1}^{q-1}b_j y^j\ne0$. Contradiction.