Is this correct? $\sum_{i_n=1}^k\sum_{i_{n-1}=1}^{i_n}...\sum_{i_3=1}^{i_4}\sum_{i_2=1}^{i_3}\sum_{i_1=1}^{i_2}i_1=\frac1{(n+1)!}\prod_{i=0}^{n-1}k+i$

$$\sum_{i_n=1}^{k}\ \sum_{i_{n-1}=1}^{i_n}\ ...\sum_{i_3=1}^{i_4}\ \sum_{i_2=1}^{i_3}\ \sum_{i_1=1}^{i_2}\ i_1 = \frac{1}{(n+1)!}\prod_{i=0}^{n-1}\ k+i $$

While solving a problem, I guessed that this above equation would be established, but I cannot prove it accurately. If this equation is correct, can we find a way to prove it? (Of course, it can be wrong.)

As written, your formula looks a bit off. In particular, take the case $n=1$; then your LHS is $\sum_{i_1=1}^k i_1=\frac{1}{2}k(k+1),$ while your RHS is $\frac{1}{2}k.$ You should be good to go if you make the product on the RHS span from $i=0$ to $i=n.$

Indeed, we have the following well-known identity:

$$\sum_{i_{n-1}=1}^{i_n}...\sum_{i_1=1}^{i_2}\sum_{i_0=1}^{i_1} 1={i_n+n-1\choose n}.$$

Geometrically, this expression can be interpreted as the number of spheres you need to stack to form an $n$-dimensional simplex with $i_n$ "rows" of spheres. For instance, the case $n=2$ produces triangular numbers, $n=3$ produces tetrahedral numbers, and so on. They appear as entries of the $n$th column (counting from $0$) of a left-justified Pascal's triangle:

The identity can be proven through a straightforward inductive combinatorial argument. We need to show for $n=0,1,2,..$ that

$$\sum_{i_{n}=1}^{i_{n+1}}{i_n+n-1\choose n}={i_{n+1}+n\choose n+1}.$$

The case $n=0$ is trivial. Now observe in general, the RHS is the number of ways $n+1$ numbers can be chosen from the set $\{1,2,...,i_{n+1}+n\}.$ Meanwhile, we can also enumerate this as

-The number of ways we can choose $n+1$ numbers having "$1$" as the smallest number (giving ${i_{n+1}+n-1\choose n}$ free choices for the other numbers), plus

-The number of ways we can choose $n+1$ numbers having "$2$" as the smallest number (giving ${i_{n+1}+n-2\choose n}$ free choices for the other numbers), plus $$\vdots $$ -The number of ways we can choose $n+1$ numbers having "$i_{n+1}$" as the smallest number (giving ${n\choose n}$ free choices for the other numbers),

all of which is the LHS.