Applying the basic identity $\frac{1}{k+1}\binom{n}{k} = \frac{1}{n+1}\binom{n+1}{k+1}$ yields: $$\begin{array} {rl} & \displaystyle \binom{n}{0}+\frac{1}{2}\binom{n}{1} + \frac{1}{3}\binom{n}{2} + \cdots + \frac{1}{n+1}\binom{n}{n}\\ \\ =&\displaystyle \frac{1}{n+1}\binom{n+1}{1} + \frac{1}{n+1}\binom{n+1}{2}+\cdots+\frac{1}{n+1}\binom{n+1}{n+1}\\ \\=&\displaystyle\frac{1}{n+1}\left(\binom{n+1}{1} + \binom{n+1}{2}+\cdots+\binom{n+1}{n+1}\right)\\ \\=&\displaystyle\frac{1}{n+1}(2^{n+1}-1) \end{array}$$