Stirling number of first kind monotone for a half 
Show that every $n>0$, there is some $m(n)$ such that $$s_{n,0}<s_{n,1}<\cdots < s_{n,m(n)}>s_{n,m(n)+1}>\cdots>s_{n,n},$$
  where either $m(n)=m(n-1)$ or $m(n)=m(n-1)+1$ and $s_{n,k}$ is the Stirling number of the first kind.

I have no clue to prove the claim. Can anyone give me some hints? Thank you in advance!
 A: SKETCH: Use the standard recurrence for the Stirling numbers of the first kind,
$${{n+1}\brack k}=n{n\brack k}+{n\brack{k-1}}\;,\tag{1}$$
to prove it by induction on $n$. (I use the notation $n\brack k$ for your $s_{n,k}$.) 
Suppose that 
$$\begin{cases}
{n\brack k}<{n\brack{k+1}},&\text{if }0\le k<m(n)\\
{n\brack k}>{n\brack{k+1}},&\text{if }m(n)\le k<n\;.
\end{cases}$$
If $1\le k<m(n)$, then ${n\brack{k-1}}<{n\brack{k}}<{n\brack{k+1}}$, and
$$\begin{align*}
{{n+1}\brack{k}}&=n{n\brack{k}}+{n\brack{k-1}}\\
&<n{n\brack{k+1}}+{n\brack{k}}\\
&={{n+1}\brack{k+1}}\;,
\end{align*}$$
so ${{n+1}\brack{k}}<{{n+1}\brack{k+1}}$, and you can easily check that this also holds for $k=0$.
Now make a similar calculation to show that ${{n+1}\brack{k}}>{{n+1}\brack{k+1}}$ when $m(n)<k<n$, and verify that the inequality also holds for $k=n$. At this point you’ll have shown that
$$\begin{cases}
{{n+1}\brack k}<{{n+1}\brack{k+1}},&\text{if }0\le k<m(n)\\
{{n+1}\brack k}>{{n+1}\brack{k+1}},&\text{if }m(n)<k<n+1\;.
\end{cases}\tag{2}$$
The only two adjacent Stirling numbers for $n+1$ whose order relationship isn’t specified by $(2)$ are ${{n+1}\brack{m(n)}}$ and ${{n+1}\brack{m(n)+1}}$, and how they’re ordered determines what $m(n+1)$ is.
