If $$(1 + x)^n = \sum_{r=0}^{n} C_r x^r,$$​ then $$\left(1 + \frac{C_1}{C_0}\right)\left(1 + \frac{C_2}{C_1}\right)\left(1 + \frac{C_3}{C_2}\right)\cdots\left(1 + \frac{C_n}{C_{n-1}}\right)$$ is equal to:

Given:
​$$(1 + x)^n = \sum_{r=0}^{n} C_r x^r \\$$​​
We are to evaluate:
​$$\left(1 + \frac{C_1}{C_0}\right)\left(1 + \frac{C_2}{C_1}\right)\left(1 + \frac{C_3}{C_2}\right) \cdots \left(1 + \frac{C_n}{C_{n-1}}\right) \\[10pt]$$​​

Concept used:
$$1 + \frac{C_r}{C_{r-1}} = \frac{C_{r-1} + C_r}{C_{r-1}} = \frac{\binom{n+1}{r}}{\binom{n}{r-1}} \quad$$​ (by Pascal's identity)
So full product becomes:
​$$\prod_{r=1}^{n} \left(1 + \frac{C_r}{C_{r-1}}\right) = \prod_{r=1}^{n} \frac{\binom{n+1}{r}}{\binom{n}{r-1}} \\$$​​

This simplifies to: $$\quad \frac{(n+1)^n}{n!} \\$$​​

Correct answer is: $${\frac{(n+1)^{n}}{n!}}$$​​