We need to show

$\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k1}}\right)$ 
$=$ 
$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)$ 

Let us begin by writing the lefthand side as

$$\frac{n!}{k!(nk)!}+\frac{n!}{(k1)!(n(k1))!}$$ 

Getting a common denominator and simplifying, we have

$\frac{n!}{k!(nk)!}}+{\displaystyle \frac{n!}{(k1)!(nk+1)!}$ 
$=$ 
$\frac{(nk+1)n!}{(nk+1)k!(nk)!}}+{\displaystyle \frac{kn!}{k(k1)!(nk+1)!}$ 



$=$ 
$\frac{(nk+1)n!+kn!}{k!(nk+1)!}$ 



$=$ 
$\frac{(n+1)n!}{k!((n+1)k)!}$ 



$=$ 
$\frac{(n+1)!}{k!((n+1)k)!}$ 



$=$ 
$\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k}}\right)$ 
