Show that the OR of n variables cannot be expressed as a polynomial over Fp of degree less than n
Let $AND:\{0,1\}^n\mapsto\{0,1\}$ be a polynomial in $GF(q), q\geq 2$.
Notice that the polynomial can't have a constant term (the constant term is zero), because
$AND(0,...,0) = 0$.
Meaning we can write the polynomial as:
$$AND(x_1,...,x_n) = \sum_{S\subseteq[1,n], S\neq \emptyset}a_S\prod_{j\in S}x_j$$
We now claim that every non-zero summand in $AND$ contains $x_i$.
Suppose that there are a set of summands not containing $x_i$. Choose the smallest (by amount of different parameters contained) of them. Notice that this  term is uniquely identified by a set $U \subset [1,n]\backslash\{i\}$. Now set $x_u = 1, u\in U$ and $x_v = 0, v\notin U$ and that there is no proper subsets of $U$ whose associated term is non-zero, because $|U|$ was chosen to be minimal.
$$AND(x_1,...,x_n) = a_U$$
This implies $a_U = 0$, because $i\notin U \land x_i = 0$. Repeat for all other summands (in increasing order by size). Thus all non-zero terms contain $x_i$. This applies to all $i\in[1,n]$, hence the degree of $AND$ is $n$.
We can denote $OR$ as $OR(x_1,...,x_n) = 1-AND(1-x_1,...,1-x_n)$ and likewise $AND(x_1,...,x_n) = 1-OR(1-x_1,...,1-x_n)$ by De Morgan's Law. Thus $AND$ and $OR$ must have the same degree.