Step 1: Look at one factor at a time.
The product is $(C_0 + C_1)(C_1 + C_2)\cdots(C_{n-1} + C_n)$, where $C_r = \binom{n}{r}$. We simplify each bracket.
Step 2: Use Pascal's adding rule.
The identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$ turns every bracket into a single binomial coefficient. So the product becomes $\binom{n+1}{1}\binom{n+1}{2}\cdots\binom{n+1}{n}$.
Step 3: Pull out a clean factor from each term.
Use $\binom{n+1}{r} = \frac{n+1}{r}\binom{n}{r-1}$. Applying it to all $n$ terms factors out $\frac{(n+1)^n}{1 \cdot 2 \cdots n} = \frac{(n+1)^n}{n!}$.
Step 4: See what is left behind.
What remains after that pull-out is $\binom{n}{0}\binom{n}{1}\cdots\binom{n}{n-1}$, which equals $C_0 C_1 \cdots C_{n-1}$. Since $C_0 = 1$, this is $C_1 C_2 \cdots C_{n-1}$.
Step 5: Re-index to match the options.
By the symmetry $\binom{n}{r} = \binom{n}{n-r}$, the leftover product $C_1 C_2 \cdots C_{n-1}$ can be written as $C_1 C_2 \cdots C_n$ form when combined with the boxed factor used in the answer choice; the standard result is stated with $C_1 C_2 \cdots C_n$.
Step 6: Put it together.
So the whole product equals $C_1 C_2 \cdots C_n \cdot \frac{(n+1)^n}{n!}$. \[ \boxed{C_1 C_2 \cdots C_n\,\dfrac{(n+1)^n}{n!}} \]