The ordinary generating function for $ζ(s)$ $$\zeta(s)^m = \sum_{n=1}^{\infty} \frac{a_n}{n^s}$$
where $ζ(s)$ is the Riemann zeta function has the ordinary generating function:
$$\sum \limits_{n=1}^{\infty} a_nx^n = x + {m \choose 1}\sum \limits_{a=2}^{\infty} x^{a} + {m \choose 2}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} + {m \choose 3}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} x^{abc} + {m \choose 4}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} \sum \limits_{c=2}^{\infty} \sum \limits_{d=2}^{\infty} x^{abcd} +... $$
$\tag1$
Wikipedia reference is http://en.wikipedia.org/wiki/Dirichlet_series
My Attempt to prove is below But I am stuck 
$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}=(1+\frac{1}{2^s}+\frac{1}{2^{2s}}+....)(1+\frac{1}{3^s}+\frac{1}{3^{2s}}+....)(1+\frac{1}{5^s}+\frac{1}{5^{2s}}+....)...=\prod_{p= prime} \frac{1}{1-p^{-s}}$$
$$\zeta(s)^m = (\sum_{n=1}^{\infty} \frac{1}{n^s})^m=(\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+.....)^m=\prod_{p= prime}(1-p^{-s})^{-m}=\prod_{p= prime}(1+mp^{-s}+\frac{m(m+1)p^{-2s}}{2!}+\frac{m(m+1)(m+2)p^{-3s}}{3!}+....)$$
If we find all prime factors for $n$
if $n=p^{b_1}_1p^{b_2}_2  p^{b_3}_3... p^{b_r}_r$  then
$$a_n=\frac{m(m+1)..(m+(b_1-1))}{b_1!}.\frac{m(m+1)..(m+(b_2-1))}{b_2!}...\frac{m(m+1)..(m+(b_r-1))}{b_r!}$$
I could not see a way after that point how to prove the generation function .
Could you please help me how to get the result shown in Equation $1$ with elementary methods?
Thanks a lot for answers
 A: 
with elementary methods

Not sure if that counts as elementary, but:
If you compute
$$\zeta(s)^m = \left(\sum_{n=1}^\infty \frac{1}{n^s}\right)^m$$
as a Cauchy product of absolutely convergent series, you can see that $a_n$ is the number of ways you can write $n$ as a product of $m$ positive integers, where order matters. $1$ is included as an allowed factor here.
Let $b_n^k$ the number of ways to write $n$ as a product of $k$ integers $> 1$, again order matters, so $12 = 2\cdot 2\cdot 3$ and $12 = 2\cdot 3 \cdot 2$ count as two different ways, hence $b_{12}^3 = 3$.
Then, for $n > 1$, you have
$$a_n = \sum_{k = 1}^m \binom{m}{k} b_n^k,$$
and $a_1 = b_1^0 = 1$. (The binomial coefficients comes from the $\binom{m}{k}$ possible choices of the factors $> 1$.)
The coefficient of $x^n$ in
$$\sum_{\substack{a_i \geqslant 2\\1 \leqslant i \leqslant k}} x^{a_1\cdot \dotsb a_k}$$
is $b_n^k$.
A: There are several processes to do this. One of those is induction method. See, $(\zeta(s))^m=(\zeta(s))^{m-1}\zeta(s)$. So if $(\zeta(s))^k=\displaystyle\sum_{n=1}^{\infty}\frac{a_k(n)}{n^s}$, Dirichlet convolution gives $a_{k+1}(n)=\displaystyle\sum_{d|n}a_k(d)$.
Now, you have $\sum \limits_{n=1}^{\infty} a_k(n)x^n = x + {m \choose 1}\sum \limits_{a=2}^{\infty} x^{a} + {m \choose 2}\sum \limits_{a=2}^{\infty} \sum \limits_{b=2}^{\infty} x^{ab} +...$. So for generating series of $(\zeta(s))^{k+1}$ will be $\sum \limits_{n=1}^{\infty} a_{k+1}(n)x^n=\sum\limits_{n=1}^{\infty}x^n\sum_{d|n}a_k(d)$ and your result followes from this.
Other thing you can do the following:
Combinatorially think if $(\zeta(s))^k=(\sum_{n=1}^{\infty}\frac{1}{n^s})^k=\sum_{n=1}^{\infty}\frac{a_k(n)}{n^s}$ where $a_k(n)=d_k(n):=\#\{(a_1, a_2,..,a_k):a_1a_2...a_k=n, a_i\in\mathbb{N}\}$.
Now, in $a_1a_2...a_k$ choose any $r$ of them and make them $1$, and rest $k-r$ will be bigger that $1$. So if $t_r(n)=\#\{(a_1, a_2,..,a_k):a_1a_2...a_k=n, a_i\geq2\}$ then $d_k(n)=\sum\limits_{r=0}^{k}{{k}\choose{r}}t_r(n)$.
So, $\displaystyle\sum_{n=1}^{\infty}a_k(n)x^n=\sum\limits_{r=0}^{k}{{k}\choose{r}}\displaystyle\sum_{n=1}^{\infty}t_r(n)x^n$.
Note that, $\sum_{n=1}^{\infty}t_r(n)x^n=\displaystyle\sum_{2\leq i_1,i_2,...,i_r \leq n}x^{i_1i_2...i_r}$. And thus your result followes.
A: You can define this OGF in terms of the polygamma function (digamma function) as follows when $s$ is a positive integer: 
$$\zeta(n) = -[z^n] \psi_0(1-z), n \in \mathbb{Z}^{+}$$
For non-integer $s$ with $\Re(s) > 1$, consider the following Taylor series expansion of the Hurwitz zeta function:
$$\zeta(s, a) = a^{-s} + \sum_{n \geq 0} \binom{s+n-1}{n} \zeta(n+s) (-a)^{n}.$$
