Answer : (A)
First let’s simplify ${n \choose k} + {n \choose k+1} = \frac{n!}{(n-k)! * k!} + \frac{n!}{(n-k-1)!*(k+1)!}$
$= \frac{(k+1)n! + (n-k)n!}{(n-k)!*(k+1)!} = \frac{n!(k+1+n-k)}{(n-k)!*(k+1)!}$
$\therefore {n\choose k} + {n \choose k+1} = \frac{(n+1)*n!}{(n-k)!*(k+1)!}$ ------------ (eq 1)
Now simplifying LHS using (eq 1)
$({n\choose 0} + {n\choose 1})({n\choose 1} + {n\choose 2}) … ({n\choose n-1} + {n\choose n}) = (\frac{(n+1)*n!}{n!*1!})(\frac{(n+1)*n!}{(n-1)!*2!})...(\frac{(n+1)*n!}{(1)!*n!})$ ------------ (eq 2)
Solving RHS
$k*{n\choose 0}*{n \choose 1}*...*{n\choose n-1} = k*(\frac{n!}{n!*0!})*(\frac{n!}{(n-1)!*1!})*(\frac{n!}{(n-2)!*2!})*...*(\frac{n!}{1!*(n-1)!})$ ------------ (eq 3)
Equating (eq 2) and (eq 3) we get –
$(\frac{(n+1)*n!}{n!*1!})(\frac{(n+1)*n!}{(n-1)!*2!})...(\frac{(n+1)*n!}{(1)!*n!}) = k*(\frac{n!}{n!*0!})*(\frac{n!}{(n-1)!*1!})*(\frac{n!}{(n-2)!*2!})*...*(\frac{n!}{1!*(n-1)!})$
Cancelling terms from LHS and RHS we get
$\frac{(n+1)^n}{n!} = k*1$
$\therefore k = \frac{(n+1)^n}{n!}$