Show that the coefficient of $x^i$ in $(1+x+\dots+x^i)^j$ is $\binom{i+j-1}{j-1}$ Show that $$\text{ The coefficient of }  x^i \text{ in } (1+x+\dots+x^i)^j \text{ is } \binom{i+j-1}{j-1}$$
I know that we have:
$\underbrace{(1+x+\dots+x^i) \cdots (1+x+\dots+x^i)}_{j\text{ times}}$
My first problem how to explain that and why -1 and not $\binom{i+j}{j}$ ? its because the $1x^0$ ?
Thank you

As far as the coefficient of $x^i$ is concerned, we might just as well look at
$$(1+x+x^2+x^3+\cdots)^j,$$
that is, at $\left(\frac{1}{1-x}\right)^j$. The Maclaurin series expansion of $(1-x)^{-j}$ has coefficient of $x^i$ equal to $\frac{j(j+1)(j+2)\cdots (j+i-1)}{i!}$. This is $\frac{(j+i-1)!}{(j-1)!i!}$, which is $\binom{j+i-1}{i}$. 
