Why does exponentiating group elements by integers always make sense? Let $G$ be a group.
For any $g\in G$, we can define a mapping $\mathbb Z\to G$ via $n\mapsto g^n$.
This map is a homomorphism between the additive group $\mathbb Z$ and $G$.
On the one hand, it's obvious why we can always do this. We just define $g^n$ as "multiply $g$ with itself $n$ times" (for $n>0$, and then straightforwardly generalise to all $n\in\mathbb Z$).
Thus, $\mathbb Z$ comes into play because we are "counting" the number of times $g$ is operating on itself.
On the other hand, this means that for any group $g\in G$, there is a homomorphism $\mathbb Z\to G$ sending $n$ to $g^n$.
Is there any reason why this group in particular, $\mathbb Z\equiv(\mathbb Z,+)$, ends up being homomorphic to subgroups of every group? Are there groups other than $\mathbb Z$ sharing this property?

Because the group is equipped with an operator which for which this set has closure. 
Anything that the operator results will remain in the set as long as the inputs are in the set.
So we can prove using induction : 


*

*Firstly, observe $g^0 = e \in G$ (base case), must exist such an $e$ (identity element)  in every group.

*Secondly because of the closure property, for any $a \in G$, we have:   $$g^{n}a = g^{n-1}\circ \underset{b\in G}{\underbrace{g \circ a}}$$
So for any $n$ this property exists iff it exists for $n-1$, and it does exist for $0$. 
Therefore by induction it must exist for all $n$.
