Here $C_k$ means $\binom{n}{k}$.
and $\binom{n}{k }$/$\binom{n}{k+1 }$=(k+1)/(n-k)
Therefore,$C_0$/$C_1$=1/(n-0) and $C_1$/$C_2$=2/(n-1), $C_2$/$C_3$=3/(n-2)...
so (1+$C_0$/$C_1$)*(1+$C_1$/$C_2$)*(1+$C_2$/$C_3$)....(1+$C_n$/$C_n-1$) =(1+1/n)*(1+2/(n-1))*(1+3/(n-2)).....(1+n/(n-n+1))
=((n+1)/n)*((n+1)/(n-1))*((n+1)/(n-2))....*((n+1)/1)
=$(n+1)^{n}$/n!.
There fore option d is the right answer
Can some one edit this... i am new to latex