Prove that $(1-\sum_{n\geq1}p_nt^n)^{-1}=\frac{\sum_{n\geq0}h_nt^n}{1-\sum_{n\geq1}(n-1)h_nt^n}$ Prove the identity:
$$(1-\sum_{n\geq1}p_nt^n)^{-1}=\frac{\sum_{n\geq0}h_nt^n}{1-\sum_{n\geq1}(n-1)h_nt^n}$$
where $h_n$ is the complete homogeneous symmetric functions for $n\in$ Par by the formula, or the sum of all monomials of degree $n$.
I know it's sort of unreasonable to ask someone to prove this identity.  But I really have no idea how to start, where to start and what other theorems I should use.
I really need this question to be answered.  A full proof would be appreciated, or a very helpful hint.  Thanks!
 A: We can use the formula:$$\prod_{i,j}(1-x_iy_j)^{-1}=\exp\sum_{n\geq1}\frac{p_n(x)p_n(y)}{n}$$
Let's make $y_1=t$, $y_2=y_3=y_4=\dots=0$, thus we get:
$$\sum_{n\geq1}p_n\frac{t^n}{n}=\log\sum_{n\geq0}h_nt^n$$
Afterwards, we can differentiate with respect to $t$, then multiply by $t$ again to get:
$$\sum_{n\geq1}p_nt^n=\frac{\sum_{n\geq0}h_nt^nn}{\sum_{n\geq0}h_nt^n}$$
which is equivalent to the desired formula.  $\square$
A: Hint:
$$\begin{array}{lrcl} & \left(1-\sum_{n=1}^\infty p_nt^n\right)\sum_{m=0}^\infty h_mt^m & = & 1-\sum_{k=1}^\infty (k-1)h_kt^k \\ \iff & \left(1-\sum \frac{x_it}{1-x_it}\right)\prod  \frac{1}{1-x_jt} & = & \left(1-t\frac{\rm d}{{\rm d}t}\right)\prod \frac{1}{1-x_jt}.\end{array}$$
Recall the power of logarithmic differentiation on products (via $\log\prod=\sum\log$):
$$\frac{{\rm d}}{{\rm d}t}\log f(t)=\frac{f'(t)}{f(t)}\iff \frac{{\rm d}f}{{\rm d}t}=f(t)\cdot\frac{{\rm d}}{{\rm d}t}\log f(t) $$
$$\iff \quad \frac{{\rm d}}{{\rm d}t}\prod f_i(t)=\left[\prod f_i(t)\right]\cdot\sum \frac{f_i'(t)}{f_i(t)}.$$
