How to prove that Mon as a category, fulfills the requirement of composition of arrows being associative? I am having trouble proving that the category Mon is a category; in particular I don't know how to prove that it fulfills $h\circ (g \circ f)=(h \circ g)\circ f$.
So the category Mon is composed of monoids, structures of the form $(M,\times,1)$, where $\times$ being a binary operator and $1$ being a unit where for all $m\in M$, $m\times 1=m=1\times m$. 
$\times$ is also associative, ie. $(m \times n) \times r = m \times (n \times r)$.
I think I already got the identity requirement: just define an arrow $f$ as the identity function which goes from $m \in M$ back to itself. Then the morphism property $f(m \times n)=m\times n=f(m)\times f(n)$ is respected.
But I don't know what to do with composition associativity. I tried plugging inputs beginning with $f$, and then try to transform it to the form I need, i.e. for $m,n\in M$, 
$h\circ (g \circ f(m\times n))$
$h\circ g \circ (f(m)\times f(n))$
$ h\circ (g(f(m))\times g(f(n))$
$h(g(f(m)))\times h(g(f(n)))$
But as you can see, this is getting me nowhere. I tried to do the same except beginning with $g$ to see if I can work back up to $h(g(f(m)))\times h(g(f(n)))$, but then using $m$ and $n$ would not make sense, because they are supposed to be members of $f$'s domain.

I am guessing this is supposed to be a ridiculously easy question, perhaps by applying definitions - but I just don't see it. 
Part of the reason maybe that I don't really understand homomorphism, which the arrows are supposed to be, on an abstract level. I've benn trying to abstract away details specific to certain domain like natural numbers or integers, but still I am going nowhere. Could anyone help please?
 A: To prove $h\circ(f\circ g)=(h\circ f)\circ g$ you do not have to work with a product $mn$. 
It is enough to observe that for every $k$ in the domain of $f$ we have:
$\begin{aligned}\left(h\circ\left(g\circ f\right)\right)\left(k\right) & =h\left(\left(g\circ f\right)\left(k\right)\right)\\
 & =h\left(g\left(f\left(k\right)\right)\right)\\
 & =\left(h\circ g\right)\left(f\left(k\right)\right)\\
 & =\left(\left(h\circ g\right)\circ f\right)\left(k\right)
\end{aligned}
$ 
It is another story to prove that the composition of two monoid-homomorphisms is again a monoid-homomorphism.
For this it is enough to observe that compositions send identities to identities (caution: this concerns the identity elements of the objects (i.e. the monoids) and not the identity arrows of the category $\mathbf{Mon}$) and respect multiplication:
$\begin{aligned}\left(g\circ f\right)\left(e\right) & =g\left(f\left(e\right)\right)\\
 & =g\left(e'\right)\\
 & =e''
\end{aligned}
$
(where $e$ denotes identity element of domain of $f$ and $e'$ denotes identity element of domain of $g$ and $e''$ identity element of codomain of $g$).
and:
$\begin{aligned}\left(g\circ f\right)\left(mn\right) & =g\left(f\left(mn\right)\right)\\
 & =g\left(f\left(m\right)f\left(n\right)\right)\\
 & =g\left(f\left(m\right)\right)g\left(f\left(n\right)\right)\\
 & =\left[\left(g\circ f\right)\left(m\right)\right]\left[\left(g\circ f\right)\left(n\right)\right]
\end{aligned}
$
It is then immediate that also $h\circ(f\circ g)=(h\circ f)\circ g$ will be a monoid-homomorphism.
If a monoid $M$ is looked at as a category itself then its identity is actually the (unique) identity arrow $M\to M$. This might cause some confusion. If you are looking at category $\mathbf{Mon}$ then do not confuse identity arrow $M\to M$ with identity element $e\in M$.
