I am having troubles with proving the above identity. Can I please ask for someone's help? I have spent more than two days on this question and I am kind of exhausted. So I would really appreciate if someone could just provide a full proof. Thanks so much for your help.
Note: In the above identity, $e_j$ denotes the elementary symmetric functions and $P_k$ denotes the power sum symmetric functions ($P_k=\alpha_1^k+\alpha_2^k+\alpha_3^k+...+\alpha_n^k$).
For any power series $f(t)=c_{1}t+c_{2}t^2+...$ with zero constant term, $exp(f(t))$ is defined by the following:
Let's see if you can get it with these two hints:
Left side: factor using Vieta's formulas.
Right side: rewrite using the Mercator series.
Alright, I'll spoil the exercise.
Left side is
$$ \begin{array}{ll} (1+tx_1)(1+tx_2)\cdots(1+tx_n) & = \phantom{+} 1 \\ & \phantom{=} +(x_1+\cdots+x_n)t \\ & \phantom{=} +(x_1x_2+\cdots+x_{n-1}x_n)t^2 \\ & \phantom{=} \qquad\qquad\quad \vdots \\ & \phantom{=} +(x_1x_2\cdots x_n)t^n. \end{array} $$
Right side is
$$ \begin{array}{l} \displaystyle \phantom{=} \exp\left(\frac{x_1+\cdots+x_n}{1} \, t \,-\, \frac{x_1^2+\cdots+x_n^2}{2} \, t^2 \,+\, \frac{x_1^3+\cdots+x_n^3}{3} \, t^3 \,-\, \cdots\right) \\[5pt] \displaystyle =\exp\left[\left(\frac{x_1t}{1}-\frac{(x_1t)^2}{2}+\frac{(x_1t)^2}{3}-\cdots\right) + \cdots + \left(\frac{x_nt}{1}-\frac{(x_nt)^2}{2}+\frac{(x_nt)^2}{3}-\cdots\right) \right] \\[5pt] \displaystyle = \exp\left(\frac{x_1t}{1}-\frac{(x_1t)^2}{2}+\frac{(x_1t)^2}{3}-\cdots\right) \,\cdots\, \exp\left(\frac{x_nt}{1}-\frac{(x_nt)^2}{2}+\frac{(x_nt)^2}{3}-\cdots\right) \\[5pt] \displaystyle =\exp\!\big(\!\ln(1+tx_1)\big) \,\cdots\, \exp\!\big(\!\ln(1+tx_n)\big) \\[5pt] \displaystyle =(1+tx_1)\cdots(1+tx_n). \end{array} $$