\centerdot 1}_{n}) =
\beta(S(\underbrace {1 \centerdot 1 \centerdot \ldots \centerdot 1}_{n})).$ Next by Theorem \[t:derivation\], $S(\underbrace {1 \centerdot 1 \centerdot \ldots \centerdot 1}_{n}) = 
n S(\underbrace {1 \centerdot 1 \centerdot \ldots \centerdot 1}_{n-1}).$ These two facts and the initial condition $\beta(1) = 1$, yield us the result $\beta(\underbrace {1 \centerdot 1 \centerdot \ldots \centerdot 1}_{n}) = n!.$

Next we record two results that will be useful in later sections. They are direct corollaries of Lemma \[l:easy\] and the fact $\beta(v) = \beta(v^*)$. However, we will not rely on this lemma. Instead, we will give independent proofs using the method illustrated in Example \[eg:warmup\]. For that, observe two simple facts, namely, $S(u)^* = S(u^*)$ and $(u \centerdot v)^* = v^* \centerdot u^*$. Also recall that the map $S$ has degree $-1$.

\[l:add\_on\] Let $u$ and $v$ be [**cd**]{}-monomials of the same degree and $w$ be any [**cd**]{}-monomial. Then we have, $$\begin{array}{r c l}
\beta(u) > \beta(v) & \mbox{\rm iff} 
         & \beta(u \centerdot w) > \beta(v \centerdot w) \\
\beta(u) = \beta(v) & \mbox{\rm iff} 
         & \beta(u \centerdot w) = \beta(v \centerdot w).
\end{array}$$

We first prove the forward implications of both statements. Since the proofs are similar, we only do the 