Let $F(n,j)$ be the number of ways to color $n$ identical balls with $j$ colors. Given the identity $$F (n, j) = \sum_{i=0}^{j-1} \binom ji F (n - j + i, j - i)$$ prove via induction that $F(n,j) = \binom{n+j-1}{j-1}$.
This was my attempt:
We proceed via strong induction. As a base case we have $n=j=1$ so that $F(n,j)=1=\binom{1+1-1}{1-1}$, so the identity holds. Inductive step for $n$: Suppose the identity holds for all $(n,j)$ pairs $(k,j)$ where $k \leq n$. We want to show that this means that the identity also holds for all pairs $(\leq n+1,j)$. We have $$F (n+1, j) = \sum_{i=0}^{j-1} \binom ji F ((n+1) - j + i, j - i)$$ and as $(n+1) - j + i \leq n$, we can use our identity then apply the inductive hypothesis to acquire
\begin{align*} F (n+1, j) &= \sum_{i=0}^{j-1} \binom ji F ((n+1) - j + i, j - i) \\ &= \sum_{i=0}^{j-1} \binom ji \binom{((n+1) - j + i) + (j-i) - 1}{(j-i)-1}\\ &= \sum_{i=0}^{j-1} \binom ji \binom{n }{j-i-1} \end{align*}
but this doesn't look very useful. Am I making a mistake somwehere?
Do you know Vandermonde's identity?
$${m+n \choose k} = \sum_{r=0}^k {m \choose r}{n \choose k-r}$$
For the combinatorial proof, consider:
How many way can we pick $k$ objects out of $m+n$, conditional on having $r$ objects out of the first $m$?
Using that, your last step is equal to ${ n + j \choose j-1 } $, which is the form that you require.