Identity $\frac { ( rn+j+1)! } { ( j+1)! {(j! × ( n-j)!)}^r}\leq { (2r+1) }^{ rn+j+1} $

Question

I am unable to prove this identity related to factorial which is to be used in an analysis question.

How to prove that $$ \frac { ( rn+j+1)! } { ( j+1)! {(j! × ( n-j)!)}^r}\leq { (2r+1) }^{ rn+j+1} $$ if $j$ belongs to $\{0,1,2,..., n\}$ and $r$ is a fixed positive integer.

Answer 1

Right Hand side: The right hand side of the expression counts in how many ways can you put $rn+j+1$ elements in $r+\color{red}{1}$ boxes, where the first $r$ boxes have two possible states for the elements, either they are colored black or white.

Left Hand side: Doing some manipulation you can check that $$\frac{(nr+j+1)!}{(j+1)!(j!\cdot (n-j)!)^r}=\frac{(nr+j+1)!}{(j+1)!(j!\cdot (n-j)!)^r}\cdot \color{blue}{\frac{(rn)!n!^r}{(rn)!n!^r}}=\color{red}{\binom{nr+j+1}{j+1}}\binom{n}{j}^r\color{magenta}{\frac{(rn)!}{n!^r}},$$ Notice that you can read this as, choosing $j+1$ objects to place in the last box( denoted by $\color{red}{red}$) and then dividing the remaining $rn$ elements into the first $r$ boxes (taking out the order inside the box, denoted by $\color{magenta}{magenta}$) and then choosing, inside every single box, $j$ elements to color them inside the box.

Together: Notice that the left hand side is just one possible arrange of the objects of the right hand side and so there should be less possibilities.