I found following statement:
Let G be a group (written multiplicatively) and let $g ∈ G$ be fixed. Then the function $f \colon\mathbb{Z}\to G$ defined by $f(n) = g^n$ is a homomorphism (laws of exponents).
As far as I know a map is a homomorphism if $f(ab)=f(a)f(b) \forall a,b$ and the law of exponents tells me that $f(nm)=a^{nm}=(a^n)^m=f(a^n)$ rather than $f(n)f(m)$...
When you are trying to show that a function is a homomorphism you need to consider what is the group operation. In this case the group operation on $\mathbb{Z}$ is regular addition, so the law of exponents tell you that: $$f(n)f(m) = g^ng^m = g^{n+m} = f(n+m)$$
So $f$ is indeed a homomorphism.