We have,
(C0+C1C0)(C1+C2C1)(C2+C3C2)(C3+C4C3)..........(Cn−1+CnCn−1)
⇒(1+C1C0)(1+C2C1)(1+C3C2)..........(1+CnCn−1)
⇒(1+n1)(1+n(n−1)n2!)..................(1+1n)
⇒(1+n1)(1+n1)(1+n1).................(1+n1)
⇒(1+n)nn!
Hence, this is the answer.