How are $1$ and $-1$ generators of the Integers? I don't understand how you can generate the entire $\mathbb{Z}$, by $1$ or $-1$ under addition
 $1^n$ for all $n\in\mathbb{Z}$ $= 1+ 1+ 1+ 1+ \ldots$. 
Then what about the negative integers?
 A: If I understand correctly, your confusion is the following: you feel like the set generated by 1 under the group law of $\mathbb Z$ (i.e. $+$) should be $\{1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, \ldots\} = \mathbb N$. This is not how the word "generate" is used in the context of groups. There are two different ways of viewing this:
The first way: the word "generate" just doesn't really mean "generate" the same way it does in the context of, say, semigroups, or logical formulas. That is, "generate" doesn't mean "apply the operations as many times as you want to these starting values". In the context of group theory, to say that a subgroup $H \subseteq G$ is generated by a subset $A \subseteq G$ just means that $H$ is the smallest subgroup of $G$ which contains $A$.
The second way: the group structure does not just consist of the group law; a group has three operations, namely the group law, the inversion, and the neutral element. This means that a subgroup $H$ generated by a set $A$ obeys not just the laws:


*

*If $a \in A$ then $a \in H$.

*If $g,h \in H$ then $g \circ h \in G$.


It also obeys the laws:


*

*$e \in H$.

*If $g \in H$, then $g^{-1} \in H$.


Thus, if we generate a group from $1 \in \mathbb Z$, we actually get, for example, the set 
$$\{
0, 1, -1, 1 + 1, -(1 + 1), 1 + 1 + 1, -(1 + 1) + -1, \ldots
\}=\mathbb Z.$$
A: When you say a group is generated by the set $S$, it means that every element of the group can be written as $g=s_1^{m_1}s_2^{m_2}\cdots s_n^{m_n}$. Now, there are two issues:


*

*This representation doesn't have to be unique.

*When $S$ is finite, there might be many different looking ways of representing the same set $S$. (again, the uniqueness can fail miserably.)


By definition, the group generated by a single element $<h>$ contains the inverse of $h$. Because, it's a group. Therefore, the subgroup generated by $1$ also contains $-1$. 
What seems to be causing the confusion in your case is that you are not paying attention to the operation of the group. Integers are considered as a group under addition. Therefore, $1^n$ is nothing but $n\times 1$. Integers under multiplication do not form a group because no number except $\{\pm 1 \}$ is invertible. I hope that it clears it up for you.
By the way, if you know linear algebra, then you know that a vector space under its vector addition operation forms an Abelian group. Then the concept of a spanning set is the same as finding a generating set for that group. If the spanning set we have chosen were linearly independent, then we'll get a unique representation for every element of the group, but without linear independence the uniqueness fails. So, for a vector space V, the group $(V,+)$ is generated by spanning sets. I thought this might be something that you'd like to know.
A: If $n\in\mathbb Z$, there are three possibilities:


*

*$n=0$: then $n\in\langle1,-1\rangle$, becaus the identity belongs to the any subgroup;

*$n>0$: then $\displaystyle n=\overbrace{1+1+\cdots+1}^{n\text{ times}}\in\langle1,-1\rangle$;

*$n<0$: then $\displaystyle n=\overbrace{(-1)+(-1)+\cdots+(-1)}^{-n\text{ times}}\in\langle1,-1\rangle$.

