Finding the elementary and power sum symmetric polynomial given complete symmetric polynomial I have been given the complete symmetric polynomial, $h_n$ as $$h_n = n, \ \ \forall n \geq 1$$
I have to show that the sequence of elementary symmetric polynomials $(e_n)$ and the power sum symmetric polynomial $(p_n)$ are periodic with period $3$ and $6$ respectively.
(Note some unknown values of $x_1,x_2,\cdots$ have been plugged into $h_n,e_n,p_n$.)
We denote the the generating functions of $h_n, \ p_n, \ e_n$ as $H(t),\ P(t),\ E(t)$ respectively, and
$$E(t)H(-t) = 1$$
$$P(t) = \frac{H'(t)}{H(t)}$$
Therefore, I get $$H(t)=\sum_{n \geq 0} h_nt^n = h_0 +\sum_{n \geq 1} nt^n = 1+\frac{t}{(1-t)^2} = \frac{1+t^2 - t}{(1-t)^2}$$
$$\implies E(t)=\frac{(1+t)^2}{1+t^2 + t}=\frac{(1-t)(1+t)^2}{1-t^3}$$
$$\implies P(t)=\frac{1+t}{(1-t)(1+t^2 - t)}=\frac{(1+t)^2}{(1-t)(1+t^3)}$$
Any kind of help will be appreciated!
Addendum: This is a question that appears in Macdonald's Symmetric Functions and Hall Polynomials, all notations are borrowed from that text. You can find an image of the original question below:

 A: What would a rational function $f(t)$ look like if its coefficients have period $2$? Well, like
$$ a+bt+at^2+bt^3+\cdots=(a+bt)(1+t^2+t^4+\cdots)=\frac{a+bt}{1-t^2}  $$
That is, a power series' coefficients have period $2$ if and only if it is a rational function expressible as a polynomial of degree $<2$ divided by $1-t^2$. Generalize to other periods.
Note it is asking to show that $(e_n)_{n\ge\color{Red}1}$ and $(p_n)_{n\ge\color{Red}1}$ are periodic, so you will need to take a look at the constant terms of the rational functions $E(t)$ and $P(t)$ before you try to determine if they are of  the appropriate form to conclude periodic power series coefficients.
A: I was able to solve the question, and I thought I should post the answer. 
Sequence of Elementary Symmetric Functions
$$E(t) = \frac{(1-t)(1+t)^2}{1-t^3} = \frac{1-t^3+t-t^2}{1-t^3}=1+\frac{t-t^2}{1-t^3}$$
$$\text{Therefore, }
e_n= \begin{cases}
1 & n = 3k+1\\
-1 & n = 3k+2\\
0 & n = 3k+3
\end{cases}
 \ \ \ \ \ \ \forall k \geq 0$$
Sequence of Power Sum Symmetric Functions
$$P(t) = \frac{(1+t)^2}{(1-t)(1+t^3)}=\frac{(1+t+t^2)(1+t)^2}{(1-t^3)(1+t^3)}=\frac{1 + 3t + 4t^2 + 3t^3 + t^4}{1-t^6}$$
$$\implies P(t) = 1+\frac{3t + 4t^2 + 3t^3 + t^4 + t^6}{1-t^6}$$
$$\text{Therefore, }
p_n= \begin{cases}
3 & n = 6k+1\\
4 & n = 6k+2\\
3 & n = 6k+3\\
1 & n = 6k+4\\
0 & n = 6k+5\\
1 & n = 6k+6
\end{cases}
 \ \ \ \ \ \ \forall k \geq 0$$
