Prove that $n\mid (n-1)!H_{n-1}$, where $n$ is an odd integer such that $n>1$ 
Show that for all odd integers $n>1$, we have:
$$n\mid (n-1)!H_{n-1}\text{, where $H_{n-1}$ is a Harmonic series}$$

Proof.
want to show exists $k_0\in\mathbb{Z}$ such that $(n-1)!H_{n-1} = nk_0$
\begin{align}
H_{n-1}
=\sum_{k=1}^{n-1}\frac{1}{k}&=\sum_{k=1}^{\frac{n-1}{2}}\frac{1}{k}+\frac{1}{n-k}\tag*{$\text{Combine first and last term}$}\\
&=\sum_{k=1}^{\frac{n-1}{2}}\frac{n}{k(n-k)}\\
&=n\sum_{k=1}^{\frac{n-1}{2}}\frac{1}{k(n-k)}\\
\end{align}
I suppose to pick:
\begin{align}
k_0
&=(n-1)!\sum_{k=1}^{\frac{n-1}{2}}\frac{1}{k(n-k)}\\
\end{align}
How do I show $k_0$ is an integer, or are there other approaches $?$
Any help would be appreciated.
 A: Case 1); $n$ is an odd prime say $p$
In this case $\frac{(p-1)!}{j}$ is an integer for $1 \leq j \leq p-1$
and for $1 \leq j < k \leq p-1$ we have 
$\frac{(p-1)!}{j} \not\equiv \frac{(p-1)!}{k}$ $mod(p)$; thus $\frac{p!}{1}, \frac{p!}{2},...,\frac{p!}{p-1}$ contains $p-1$ distinct elements $mod(p)$ each not divisible by $p$; hence
$\frac{(p-1)!}{1}, \frac{(p-1)!}{2},...,\frac{(p-1)!}{p-1}$ is equivalent to the list $1,2,...,p-1$ $mod(p)$;
Hence we have  
$\sum_{j=1}^{p-1}\frac{(p-1)!}{j} \equiv \sum_{j=1}^{p-1}j \equiv 0$ $mod(p)$
Case 2); $n$ is an odd composite number
note that for $1 \leq j \leq n-1$ we have that $\frac{(n-1)!}{j!}$ is an integer.
Mini-case (i); There exists a prime $q$ such that $q | n$ and highest power of $q$ less than or equal to $n-1$ is $q$ (i.e $q^2 \geq n$)
Define $ord_q(n)$ to be the exponent of the highest power of $q$ dividing $n$ ; one has $ord_q(n) \leq 2$; 
if $ord_q(n) = 1$; then note 
$ord_q((n-1)!) = \frac{n}{q} - 1 \geq 2$; hence for $1 \leq j \leq n-1$ we have $ord_q(\frac{(n-1)!}{j}) \geq 1$ and thus $q | (n-1)!H_{n-1}$
If $ord_q(n) = 2$; then note $n=q^2$ and $ord_q(q^2-1) = q - 1$; If $q-1 > 2$ then for $1 \leq j \leq n-1$ $ord_q(\frac{q^2-1}{j}) \geq 2$; thus $q^2| (q^2-1)!H_{q^2-1}$. If $q-1 = 2$ then $q = 3$ and it can be checked that for $n = 3^2 = 9$ that the result is true.
Mini-case (ii) There exists a prime $q$ such that $q | n$ and highest power of $q$ less than or equal to $n-1$ is $q^h$ with $h \geq 2$.
In this mini-case we have $ord_q(n) \leq h+1$ and $ord_q((n-1)!) \geq q^{h-1}+q^{h-2}+...+1$. For $1 \leq j \leq p-1$ we have $ord_q(\frac{(n-1)!}{j}) \geq q^{h-1}+q^{h-2}+...+1 - h \geq h+1$ for $q \geq 5$ or $h \geq 3$; as $ord_q(n) \leq h+1$ we have 
$q^{ord_q(n)}|(n-1)!H_{n-1}$ for $q \geq 5$ or $h \geq 3$. For $q = 3$ and $h = 2$ we need to test all $n$ multiple of $3$ and odd in the interval $[10,26)$ and check that the result is true.
By uniqueness of prime factorization we are done.
