Solution:
By making use of the formula,
$ P\text{ }\left( n,\text{ }r \right)\text{ }=\text{ }n!/\left( n-r \right)! $
$ P\text{ }\left( n,\text{ }n \right)\text{ }=\text{ }n!/\left( n-n \right)! $
$ =\text{ }n!/0! $
$ =\text{ }n!\text{ }\left[ Since,\text{ }0!\text{ }=\text{ }1 \right] $
Let’s take the LHS:
$ =\text{ }1.\text{ }P\left( 1,\text{ }1 \right)\text{ }+\text{ }2.\text{ }P\left( 2,\text{ }2 \right)\text{ }+\text{ }3.\text{ }P\left( 3,\text{ }3 \right)\text{ }+\text{ }\ldots \text{ }+\text{ }n\text{ }.\text{ }P\left( n,\text{ }n \right) $
$ =\text{ }1.1!\text{ }+\text{ }2.2!\text{ }+\text{ }3.3!\text{ }+\ldots \ldots \ldots +\text{ }n.n!\text{ }\left[ Since,~P\left( n,\text{ }n \right)\text{ }=\text{ }n! \right] $
$ =\sum\limits_{r=1}^{n}{r.r!} $
$ =\sum\limits_{r=1}^{n}{r.r!}+r!-r! $
$ =\sum\limits_{r=1}^{n}{(r+1)r!-r!} $
$ =\sum\limits_{r=1}^{n}{(r+1)-r!} $
$ =\text{ }\left( 2!-1! \right)\text{ }+\text{ }\left( 3!-2! \right)\text{ }+\text{ }\left( 4!-3! \right)\text{ }+\text{ }\ldots \ldots \ldots \text{ }+\text{ }\left( n!-\left( n-1 \right)! \right)\text{ }+\text{ }\left( \left( n+1 \right)!-n! \right) $
$ =\text{ }2!-1!+3!-2!+4!-3!+\ldots \ldots +n!-\left( n-1 \right)!+\left( n+1 \right)!-n! $
$ =\text{ }\left( n+1 \right)!-1! $
$ =\text{ }\left( n\text{ }+\text{ }1 \right)!-1\text{ }\left[ Since,~P\text{ }\left( n,\text{ }n \right)\text{ }=\text{ }n! \right] $
$ =\text{ }P\left( n+1,\text{ }n+1 \right)-1 $
$ =\text{ }RHS $
Hence Proved.