$\text{ If }(1+\mathrm{x}{)}^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}} {\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{r}}$ then $\left(1+\frac{{\mathrm{C}}_1}{{\mathrm{C}}_0}\right)\left(1+\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}\right)\cdots \left(1+\frac{{\mathrm{C}}_{\mathrm{n}}}{{\mathrm{C}}_{\mathrm{n}-1}}\right)$ is equal to

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