If we are considering a field with characteristic $p > 0$, then it implies that the sum of $p$ 1's is 0. Therefore, $1+1+...+1 = p = 0$. However, when considering $p$ as an exponent, we do not take $p$ to be 0. For example, in $\mathbb{Z_5}$, we have that $2^5 = 32 = 2 \neq 1 = 2^0$. So, $2^5 \neq 2^0$. Why does this not imply that $5 \neq 0$ in the field $\mathbb{Z_5}$?
I'm confused as to why in a field of characteristic $p > 0$, we cannot take $p$ to be 0 in the exponent of a term. I am assuming I am missing something rather simple, but would appreciate if someone could point out what it is I'm overlooking.
When you write $2^5$ while working in field $\mathbb{Z_5}$, the symbol $2 \in \mathbb{Z_5}$, but the "$5$" is not an element of the field. It simply a way to avoid writing $2(2)(2)(2)(2)$.
Also, continuing to talk about $p=5$, remember that the elements are most simply labelled as $\mathbb{Z_5}\,=\,\{0,1,2,3,4\}.$ In which case the following is no good:
$$1+1+1+1+1\,=\,5.$$
Rather I'd put:
$$\underbrace{1+1+1+1+1}_\text{$p$ copies}\,=\,0. $$