In Paul Goerss and John Jardine's book on simplicial homotopy theory they define
the nerve $BC$ of a category $C$ as simplicial set determined by $BC([n])=hom_{cat}([n],C)$.
They then say in other words an $n$- simplex (by definition an element of $hom_{cat}([n],C)$) is a string of composable arrows of length $n$ in $C$.
I don't thing these are equivalent because there are more strings of composable arrows than there are elements in the hom-set:
Take for instance $C=\mathbb{Z}/5$ where $\mathbb{Z}/5$ is the category with one object and one morphism for each element in the group $\mathbb{Z}/5$. Every functor $F$ from the category $[n]$ to the category $\mathbb{Z}/5$ is determined by what $F(0\leq 1 \leq 2... \leq n)$ is. $F(0 \leq 1 \leq 2 ... \leq n )$ can be one of $5$ different arrows in the category $\mathbb{Z}/5$. Therefore there are $5$ different elements in $hom_{cat}([n], \mathbb{Z}/5)$.
The number of composable arrows of length $n$ is $5^n$.
What is wrong?
This is not true. The morphism $0\leq n$ is just one morphism in the category $[n]$, which does not generate the entire category (if $n>1$). For instance, knowing what $F(0\leq n)$ is does not determine what $F(0\leq 1)$ is.
Knowing where $F$ sends the entire diagram $0\leq 1\leq 2\leq\dots\leq n$ does determine $F$, but that diagram does not consist of just a single map: it has a map $0\to 1$, a map $1\to 2$, and so on. These are exactly a string of $n$ composable arrows, and so give a bijection between such functors $F$ and strings of $n$ composable arrows.