How to prove this symmetric polynomial identity According to this (equation 21 and 26) how to show that
\begin{gather}
\ln(1+\Pi_1t+\Pi_2t^2+\Pi_3t^3+...) =\Pi_1t+\frac{1}{2}(-\Pi^2_1+2\Pi_2)t^2+\frac{1}{3}(\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3)t^3+.. .=\sum_{k=1}^{\infty}\frac{s_k}{k}t^k 
\end{gather}
$s_1=\Pi_1 \\ s_2=-\Pi^2_1+2\Pi_2 \\ s_3=\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3 $ 
And, I've been curious about how did they derive this equation
$(-1)^{p-1}s_p = $
\begin{vmatrix}
    \Pi_1 & 1 & 0 & 0 &\dots  & 0 \\
    2\Pi_2 & \Pi_1 & 1& 0 &\ddots &0 \\
    3\Pi_3 & \Pi_2 & \Pi_1 & 1 &\ddots & 0 \\
    4\Pi_4 & \Pi_3 &\Pi_2 &\Pi_1 &\ddots&0 \\
    \vdots & \vdots & \vdots & \vdots & \ddots &1 \\
    p\Pi_{p} & p\Pi_{p-1} & p\Pi_{p-2} & p\Pi_{p-3}  & \dots & \Pi_1
\end{vmatrix}
Is there any elementary way to show? Thanks in advance
 A: Given a generating function
$$ f(x) := 1+a_1t+a_2t^2+a_3t^3+\dots \tag{1} $$
whose logarithm is $$ \ln(f(x)) =
\sum_{k=1}^{\infty}\frac{s_k}{k}t^k, \tag{2} $$
differentiate both sides to get
$$ \frac{d}{dx} \ln(f(x)) = f'(x)/f(x) =
\sum_{k=1}^{\infty}s_k t^{k-1}. \tag{3}  $$
Multiply both sides by $\,f(x)\,$ to get
$$ f'(x) = \sum_{k=1}^\infty k\,a_k\,t^{k-1} =
\left(1 + \sum_{k=1}^\infty a_k\,t^k\right)
\left(\sum_{k=1}^{\infty}s_k t^{k-1}\right). \tag{4} $$
For each $\,n>0\,$ this gives a system of equations linear in $\,s_1,s_2,\dots,s_n\,$ with coefficients using $\,a_1,a_2,\dots,a_n.\,$
Use Cramer's rule
to solve for $\,s_n\,$ which leads to the MathWorld determinant
$$(-1)^{n-1}s_n \!=\! 
\begin{vmatrix}
1\,a_1 & 1 & 0 & 0 &\dots  & 0 \\
2\,a_2 & a_1 & 1& 0 &\ddots & 0 \\
3\,a_3 & a_2 & a_1 & 1 &\ddots & 0 \\
4\,a_4 & a_3 & a_2 & a_1 &\ddots& 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots &1 \\
n\,a_n & a_{n-1} & a_{n-2} & a_{n-3} & \dots & a_1
\end{vmatrix}. \tag{5} $$
Here is an explicit example. Let $\,n=3.\,$ The linear equations are
$$ 1s_1 \!=\! 1a_1,\;\; a_1s_1 \!+\! 1s_2 \!=\! 2a_2,
\;\; a_2s_1 \!+\! a_1s_2 \!+\! 1s_3\!=\! 3a_3. \tag{6} $$
Setting this up as a matrix equation we get
$$ \begin{bmatrix}1 & 0 & 0 \\a_1 & 1 & 0\\
a_2 & a_1 & 1 \end{bmatrix} \begin{bmatrix} 
s_1 \\ s_2 \\ s_3 \end{bmatrix} =
\begin{bmatrix} 
1\,a_1 \\ 2\,a_2 \\ 3\,a_3 \end{bmatrix}\!. \tag{7} $$
Cramer's rule solving for $\,s_3\,$ and using the
denominator $3\times 3$ determinant $=1$ gives
the solution in equation $(5)$ for $\,n=3\,$ and that is
$$ s_3 = a_1^3 -3a_1a_2 + 3a_3 = (-1)^{3-1}s_3 = \begin{vmatrix}
1\,a_1 & 1 & 0 \\ 2\,a_2 & a_1 & 1 \\
3\,a_3 & a_2 & a_1 \end{vmatrix}. \tag{8} $$
