等差数列多层嵌套求和等式的证明问询

Hey everyone, let's dive into this proof problem involving nested sums of arithmetic progressions. Here's the core statement we need to verify:

Given ${a_n}$ is an arithmetic progression, prove that:
$$\sum_{x_n=1}^{{x_{n+1}}{\sum_{x_{n-1}=1}}{x_{n}}{\dots\sum_{x_1=1}^{x_2}{a_{x_1}}}} = \frac{\prod_{i=0}^{n-1}(x_{n+1}+i)}{n!}\cdot\frac{na_1+a_{x_{n+1}}}{n+1}$$

Example Illustrations #

To make this concrete, here are the cases for small values of $n$:
$$\begin{align*}
n&=1: &\sum_{x_1=1}^{x_2}{a_{x_1}} &= a_1+a_2+\dots +a_{x_2}= x_2\cdot\frac{a_1+a_{x_2}}{2}\
n&=2: &\sum_{x_2=1}^{{x_3}\left(\sum_{x_1=1}}{x_2}{a_{x_1}}\right) &= \sum_{x_2=1}^{x_3}(a_1+a_2+\dots +a_{x_2})\
&&&= (a_1)+(a_1+a_2)+\dots+(a_1+a_2+\dots +a_{x_3})\
&&&= \frac{x_3(x_3+1)}{2!}\cdot\frac{2a_1+a_{x_3}}{3}\
\end{align*}$$

Equivalent Formulation #

This problem can also be phrased using the following definition:
$$\text{Let }\ C_k = \frac{\prod_{i=0}^{k-2}(x_k+i)}{(k-1)!}$$
Then we need to prove:
$$\sum_{x_n=1}^{x_{n+1}}{C_n\cdot\frac{(n-1)a_1+a_{x_n}}{n}} = C_{n+1}\cdot\frac{na_1+a_{x_{n+1}}}{\dots}$$

备注：内容来源于stack exchange，提问作者user1317348