On a diagram like that you would use $n_1=n, n_2=n, \ldots, n_k=n, \ldots, n_r= n$
There would be $r$ distinct steps each with $n$ possible ways. That corresponds to selecting one of $n$ boxes to put a ball into, for $r$ balls.
The diagram would consist of $r$ lines, the starting point branching to $n$ points on the first line, each one with $n$ branches to the the next line, and each of those $n^2$ points starting $n$ branches to the third line, and each of those $n^3$ points branching $n$ times ... until each of the $n^{r-1}$ points on the penultimate line start $n$ branches to the last line, for a total of $n^r$ points on line $r$.
$$\underbrace{n_1\times n_2 \times \cdots \times n_r}_{\text{a term for each of $r$ balls}} = \underbrace{\quad n\times n\times \cdots \times n\quad}_{\text{each term is the count of boxes, $n$}} = n^r$$
$r$ identical balls in $n$ distinct boxes, $r\ge n$, with at least one ball in each box is equivalent to
$$[(r-n) \text{identical balls in }n\text{ distinct boxes}]\cdot[n\text{ identical balls in }n\text{ distinct boxes}].$$
The problem separates into two regions where the total number of balls in each region is fixed, AND we are only concerned with rearrangements within each region. The latter holds because the balls are identical - any swap between the two regions (swaps of course maintaining the constraint of total ball number in each region) will not generate a new solution.
The "$n$ identical balls in $n$ boxes" region is a question of permutations of the $n$ balls. This is $\dfrac{n!}{n!}=1$ - not just $n!$. Because all the balls are identical, all the rearrangements are also identical.
$R$ identical balls in $N$ distinct boxes is given by $C(R+N-1,N-1)$ - considering $N-1$ "separators" + $R$ balls, the problem is reduced to counting permutations e.g. $4$ boxes $5$ balls ~ number of permutations of $XXXxxxxx$ where the $X$ delimit the boxes.
The solution is then $C(r-n+n-1,n-1)$, as stated.
Best Answer
Let $a_k(m,n)$ be the number of ways of placing $n$ distinct objects in $k$ distinct boxes if there must be at least $m$ objects in each box. Suppose first that $k=1$. Clearly $a_1(m,n)=0$ if $n<m$, and $a_1(m,n)=1$ if $m\ge n$, so the exponential generating function for the $a_1(m,n)$ is $$A_1(x)=\sum_{n\ge 0}a_1(m,n)\frac{x^n}{n!}=\sum_{n\ge m}\frac{x^n}{n!}=e^x-\sum_{n=0}^{m-1}\frac{x^n}{n!}.$$
Suppose now that you know $A_k(x)$ for some $k\ge 1$. To distribute $n$ objects amongst $k+1$ boxes you must first choose some of them to go into box $k+1$ and then distribute the remainder amongst the first $k$ boxes. Suppose that you put $r$ objects into box $k+1$. You can choose these $r$ objects in $\binom{n}r$ ways, and there are then $a_1(m,r)$ ways to ‘distribute’ them to box $k+1$ and $a_k(m,n-r)$ ways to distribute the remainder amongst the other $k$ boxes. Summing over the possible values of $r$, we find that $$a_{k+1}(m,n) = \sum_{r=0}^n\binom{n}r a_1(m,r)a_k(m,n-r)\;,$$ which is exactly what is needed to give us $A_{k+1}(x)=A_1(x)A_k(x)$. By induction, then, $$A_k(x)=A_1(x)^k=\left(e^x-\sum_{n=0}^{m-1}\frac{x^n}{n!}\right)^k\tag{1}.$$ If we set $$p_m(x)=\sum_{n=0}^{m-1}\frac{x^n}{n!},$$ the Taylor polynomial of order $m-1$ for the exponential function, $(1)$ can be written simply $$A_k(x) = \big(e^x-p_m(x)\big)^k.$$