Suppose $$H(t)=1+\sum_{n=1}^{\infty}h_nt^n $$ is a formal power series with $h_n$ belonging to a sufficiently nice algebra. I am puzzled by the following. I am reading a paper and the author claims we can take the log of the series $$ \ln(H(t))$$ and furthermore we can write it as a power series $$\ln(H(t))=\sum_{m=1}^{\infty} \frac{P_m}{m}t^m $$ The justification is that we can identify $h_n$ the $n$th complete symmetric functions in a large number of formal variable $x_1,x_2,\ldots,x_N$. In which case $H(t)$ can be written as $$H(t)=\prod_{i=1}^N\frac{1}{1-x_it}. $$ Now if you take the log of this product we have $$\ln(H(t))=\ln(\prod_{i=1}^N\frac{1}{1-x_it})=-\sum_{i=1}^N \ln(1-x_it) $$ Using the standard power series expansion for $\ln(1-z)$ we obtain $$\ln(H(t))=\sum_{i=1}^N \sum_{k=0}^{\infty}\frac{x_i^kt^k}{k}=\sum_{k=1}^{\infty} \frac{P_k}{k}t^k$$ where $P_k=\sum_{j=1}^N x_i^k$ are the normal power sums. It appears that the original claim that $$\ln(H(t))=\sum_{m=1}^{\infty} \frac{P_m}{m}t^m $$ is justified assuming you can actually identify $h_n$ with the $n$th complete symmetric polynomial.
So my question is this: What allows one to identify $h_n$ with the $n$th complete symmetric polynomial???