Let $P_k$ be a degree-$k$ real polynomial of $m$ variables $x_1,\cdots,x_m$. $P_k$ can be written as $$P_k(x_1,\cdots,x_m) = \sum_{\alpha_1+\cdots+\alpha_m\leq k} \xi_{\alpha_1,\cdots,\alpha_m}x_1^{\alpha_1}\cdots x_m^{\alpha_m}.$$
My question is to prove the following:
$P_k(x_1,\cdots,x_k)=0$ almost everywhere on $0\leq x_1,\cdots,x_k\leq 1$ if and only if $\xi_{\alpha_1,\cdots,\alpha_m}=0$ for all monomials $\alpha_1,\cdots,\alpha_m$.
The "if" part is trivially true. Also, in univariate case ($m=1$) one can extract the non-zero components and use the fundamental theorem of algebra. However, I cannot think of a simple proof that handls $m>1$.
Over $\mathbb R$: take $Q_k=P_k^2$ and then $Q_k(x_1,\ldots,x_k)=0$ a.e. on $[0,1]^k$. This means that
$$\int_{[0,1]^k} Q_k(x_1,\ldots,x_k)dx_1\cdots dx_n=0$$
Having in mind that $Q_k$ is continuous and nonnegative, it follows that $Q_k(x_1,\ldots,x_k)=0$ on $[0,1]^k$, so also $P_k(x_1,\ldots,x_k)=0$.
All we need to prove now is that all coefficients are zero, but notice that, up to a constant factor, they are (or, at least, the degree-$k$ ones are) partial derivations of $P_k$ of some sort, thus they must also all be 0.
To see that on an example, look at e.g. $P_3(x,y)=ax^2y+\ldots$, then e.g. $a=\frac12\frac{\partial^3P}{\partial x^2\partial y}$ The general statement can be proven by induction on $k$.
Note: over $\mathbb C$, you would take $Q_k=P_k\overline{P_k}$ and the rest of the proof is the same.