I have found a interesting thing but I cannot prove it. Given $k_i$ are positive for any $i\geq1$, and we have $M+1$ by $M+1$ matrix $A$, which is $ A=\left[\begin{array}{ccccc} 0\\ k_{1} & 0\\ k_{2} & \frac{1}{2}k_{1} & 0\\ \vdots & & & \ddots\\ k_{M} & \frac{M-1}{M}k_{M-1} & \cdots & \frac{1}{M}k_{1} & 0 \end{array}\right] $ I found that the first column of $n!A^{n}$ is always equal with the first column of $B^{n}$ for any $n\in\mathbb{N}$, where $ B=\left[\begin{array}{ccccc} 0\\ k_{1} & 0\\ k_{2} & k_{1} & 0\\ \vdots & & & \ddots\\ k_{M} & k_{M-1} & \cdots & k_{1} & 0 \end{array}\right] $ Can you help me to prove it? Thanks in advance.
The following is a reply to a comment from @GerryMyerson. (Updated on Aug. 27.)
To @GerryMyerson. Here is the problem when I use the induction:
Use induction method. First, it is obvious that when $n=0$ and $1$, the equality is hold for any $M\in\mathbb{N}$. Then, we assume when $n=N>1$, the equality is hold, and the matrix is $ N!A_{M+1}^{N}=B_{M+1}^{N}=\left[\begin{array}{ccccccc} 0\\ \vdots\\ 0\\ t_{1} & 0\\ t_{2} & s & 0\\ \vdots & \vdots & & \ddots\\ t_{M+1-N} & u & \cdots & v & 0 & \cdots & 0 \end{array}\right]. $ Here, we only focus on the first column.
When $n=N+1$, the left side is $ \left(N+1\right)A_{M+1}\times\left(N!A_{M+1}^{N}\right)=\left(N+1\right)\left[\begin{array}{ccccccc} 0\\ k_{1} & 0\\ k_{2} & \frac{1}{2}k_{1} & 0\\ \vdots & & & \ddots\\ k_{i} & \frac{i-1}{i}k_{i-1} & \cdots & \frac{1}{i}k_{1} & 0\\ \vdots & & & & & \ddots\\ k_{M} & & & & & \frac{1}{M}k_{1} & 0 \end{array}\right]\left[\begin{array}{ccccccc} 0\\ \vdots\\ 0\\ t_{1} & 0\\ t_{2} & s & 0\\ \vdots & \vdots & & \ddots\\ t_{M+1-N} & u & \cdots & v & 0 & \cdots & 0 \end{array}\right], $ then the $\left(i+1,1\right)$ entry is $\left(N+1\right)\left(\frac{i-N}{i}k_{i-n}t_{1}+\frac{i-N-1}{i}k_{i-N-1}t_{2}+\cdots+\frac{1}{i}k_{1}t_{i-N}\right)$.
On the other side, the $\left(i+1,1\right)$ of $B$ entry is $k_{i-N}t_{1}+k_{i-N-1}t_{2}+\cdots k_{1}t_{i-N}$.
Now, I cannot prove that these two are equal.