A finite sum of $1$ equals $0$ in a field with finitely many elements. 
I need to prove that for field $\mathbb{F}$ with finitely many elements $\exists n \in \mathbb{N}$, $\underbrace{1+1+1+...+1}_{n \text{ times }} = 0$.

I can see why this is true, since there is a finite number of elements. A field with elements {$0,1$}, $1+1 = 0$, so $n = 2$. For set {$0,1,2$}, $1+1+1=0$, so $n$ is equal to the cardinality of the set.
I can't however, see how to prove this Mathematically.
What I want to know: Can I prove this via the field axioms solely?
If my title is poorly labeled, please edit, or inform me to edit!
Thanks for any help!
 A: A field is a group with respect to $+$ and it's identity is $0$. in a group the order of any element divides the order of group (number of element).
A: To add to what has already been said suppose $p,q\in \mathbb N$ and that $pq\cdot 1=0$ (where this means adding $pq$ ones together), then $(p\cdot 1)(q\cdot 1)=0$ so that either $p\cdot 1=0$ or $q\cdot 1=0$.
Hence in a field the least $n$ for which $n\cdot 1=0$ is a prime (in an infinite field like $\mathbb R$ there may be no such $n$).
A: The usual trick with finite structures is that when you produce a sequence of things (e.g. by repeatedly adding $1$), that eventually you have to have a repeat.
If the same field element appears twice on the list, what can you say about that? (hint: convert this fact into an equation)
As an aside, $n$ can be smaller than the cardinality of the field. e.g. the finite field with 9 elements has characteristic $3$. You can construct the finite field with 9 elements by adding $\mathbf{i}$ -- i.e. a square root of $-1$. Then the set of elements $(a+b \mathbf{i})$ with $a,b \in \mathbf{F}_3$ turns out to be a field; the proof is pretty much the same as proving the complex numbers (constructed in the same manner but starting from the reals) are a field.
A: Let $1$ be the multiplicative identity of the finite field $\mathbb F$ where $|\mathbb F| = q$.  Now, consider the following list of field
elements:
$$1\\
1+1\\
1+1+1\\
1+1+1+1\\
\cdots\\
\underbrace{1 + 1 + \cdots + 1}_{q+1 ~\text{ones}}
$$
(I refuse to call these elements $1, 2, 3, 4$ etc.) Since the field has 
$q$ elements, these $q+1$ field elements that we have listed above cannot 
all be distinct elements of the field: at least
two elements in the list must be the same. Let $\underbrace{1+1+\cdots + 1}_{i~\text{ones}}$ be an element that is a repeat of an element that
has been listed previously: that is,
$$\underbrace{1+1+\cdots + 1}_{i~\text{ones}} 
= \underbrace{1+1+\cdots + 1}_{j~\text{ones}} ~\text{for some} ~ j < i.$$
Thus, we have that
$$\begin{align}
\underbrace{1+1+\cdots + 1}_{j~\text{ones}} 
= \underbrace{1+1+\cdots + 1}_{i~\text{ones}}
=  (\underbrace{1+1+\cdots + 1}_{j~\text{ones}}) + (\underbrace{1+1+\cdots + 1}_{i-j~\text{ones}})
\end{align}$$ 
It follows that $\underbrace{1+1+\cdots + 1}_{i-j~\text{ones}} = 0$.
Note that $i$ might be as large as $q+1$ but since $j \geq 1$, we
conclude that 

there exists a positive integer $n \leq q = |\mathbb F|$ such that
  $$\underbrace{1+1+\cdots + 1}_{n~\text{ones}} = 0$$

Note that it is not claimed that $n$ is the smallest such
integer or that $n$ is a prime, just a proof that a sum of $n$ copies
of the multiplicative identity equals $0$.
