I am using the following characterization of the simplex category $\Delta$:
The objects are finite ordinals and the arrows are weakly monotone functions $f:n\rightarrow m$. $0$ is initial and $1$ is terminal in $\Delta $, so we may define $\mu ^{\left ( k \right )}:k\rightarrow 1$, recursively by taking
$\mu ^{\left ( 0 \right )}=\eta :0\rightarrow 1$; $\mu ^{\left ( 1 \right )}=1:1\rightarrow 1$; $\mu =\mu ^{\left ( 2 \right )}:2\rightarrow 1$; $\mu ^{\left ( 3 \right )}=\mu (\mu+1)=\mu (1+\mu ):3\rightarrow 1$ $\mu ^{\left ( 4 \right )}=\mu ((\mu+1) +1)):4\rightarrow 1$
Then, define a bifunctor $+:\Delta \times \Delta \rightarrow \Delta$ by the usual $+$ on the objects and for arrows
$f:n\rightarrow m; g:n'\rightarrow m'$ by
$(f+g)(i)=f(i)$ if $i=0, 1, ...n-1$ $(f+g)(i)=m+g(i-n)$ if $i=n, n+1, ..., n+n'-1$.
Then $\left \langle \Delta ,+, 0\right \rangle $ is a monoidal category and $\left \langle 1 ,\mu, \eta \right \rangle$ is a monoid in $\Delta $.I want to prove that if $f:n\rightarrow m$, then $f$ is a sum of such iterated products (in $\mu $ and $\eta )$.
All I have so far is, by using the definition of $\mu^{n}$, the associative law
$\mu ^{n}(\mu ^k_{1}+...+\mu ^k_{n})=\mu^{k_{1}+....+k_{n}}$. I am not sure how to proceed from here.
Edit: I was playing around with this and I just took a simple case: $f:4\rightarrow 3$ defined by $f(0)=f(1)=f(2)=1$ and $f(3)=2$. Then I considered
$f^{-1}\left ( 0 \right )=\emptyset \\f^{-1}\left ( 1 \right )=\left \{ 0,1,2 \right \} \\f^{-1}\left ( 2 \right )=\left \{ 1 \right \} $.
Then when I calculate $\mu ^{\left ( 0 \right )}+\mu ^{\left ( 3 \right )}+\mu ^{\left ( 1 \right )}$ and it is equal to $f$. So it appears that if I take $f^{-1}\left ( j \right )$ for each $j\in m$, and use the ordinal that it corresponds to, then in fact, $f$ is a sum of the required arrows.
To prove it in general, I let $n_{j}=f^{-1}\left ( j \right )$ for $0\leq j\leq m$ . $n_{j}$ is the ordinal that corresponds to the number of elements in $n$ that $f$ sends to $j\in m$. Then $\sum_{j=0}^{m}n_{j}=n$.
Then consider $f_{0}=\mu ^{n_{0}}+\mu ^{n_{1}}:n_{0}+n_{1}\rightarrow 1 $. By definition of $+$, $f_{0}\left ( i \right )=0$ if $0\leq i\leq n_{0}-1$ and $f_{0}\left ( i \right )=1$ if $n_{0}\leq i\leq n_{0}+n_{1}-1$.
Next, consider $f_{1}=f_{0}+\mu ^{n_{2}}:(n_{0}+n_{1})+n_{2}\rightarrow 1 $. Again by definition of $+$, we have $f_{1}(i)=f_{0}(i)$ if $0\leq i\leq n_{0}+n_{1}-1$ and $f_{1}(i)=2$ if $n_{0}+n_{1}\leq i\leq n_{0}+n_{1}+n_{2}-1$.
I continue in this way, setting $f_{j}=f_{j-1}+\mu ^{n_{j+1}}$.
Now, $f(0)$ is some $j\in m$ so $m_{k}=0$ for $0\leq k< j$ because $f$ is monotone. Clearly, again by definition of $+$, $(\mu ^{m_{0}}+\cdot \cdot \cdot +\mu ^{m_{j}})(0)=j$ because all the terms previous to $\mu ^{m_{j}}$ are $\mu ^{0}$ and in fact $(\mu ^{m_{0}}+\cdot \cdot \cdot \mu ^{m_{j}}+\cdot \cdot \cdot +\mu ^{n_{m}})(0)=j$ as well because the $\mu $-terms tacked on do not affect the value of the sum at $0$. So the result is true if $i=0$.
Likewise, $f(1) =j_{1}\geq j$ and so $(\mu ^{m_{0}}+\cdot \cdot \cdot +\mu ^{m_{j}}+\cdot \cdot \cdot +\mu ^{n_{j_{1}}})(1)=j_{1}$, and arguing as before, the whole sum mut be $j_{1}$.
Following this recipe, we obtain $f$ as the required sum.
Please comment on this proof. I'd like to see it made cleaner.