Hypothesizing closed form of $(\sum_{n=0}^{\infty} x^n)^i$, where $i\ge 1$, $i$ is an integer

In my previous question Question

We have arrived at the conclusion that $$(\sum_{n=0}^\infty x^n)^2=\sum_{n=0}^\infty (n+1)x^n$$

And I found this infinite series can be written as $$\sum_{i=0}\binom{n+2-1}{2-1}x^n$$

While that it is of the form $(\sum_{i=0}^{\infty} x^i)^j$, we could also find that $$(\sum_{n=0}^\infty x^n)^1=\sum_{n=0}^\infty\binom{n+1-1}{1-1}x^n$$

And by calculating the case for $j=3$

And enumerate the coefficient for $n$ which is the number of non-negative integral solutions to $i+j+k=n$, by fixing one variable $i$, and counting on the cases it is $\frac{(n+2)(n+1)}{2}=\binom{n+3-1}{3-1}$

So we can hypothesize by generalizing that $$(\sum_{n=0}^\infty x^n)^i=\sum_{n=0}^{\infty}\binom{n+i-1}{i-1}x^n$$ for $i\ge 1, i\in\mathbb{N}$

This serves as a complement to my previous question.