How to prove $\sum\limits_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$? Other than the general inductive method,how could we show that $$\sum_{r=0}^n \frac{(-1)^r}{r+1}\binom{n}{r} = \frac1{n+1}$$
Apart from induction, I tried with Wolfram Alpha to check the validity, but I can't think of an easy (manual) alternative.
Please suggest an intuitive/easy method.

Note that
$$\begin{align*}\frac{(n+1)\binom{n}{r}}{r+1} &= \frac{(n+1)n!}{(r+1)r!(n-r)!} = \frac{(n+1)!}{(r+1)!(n-r)!}\\ &= \frac{(n+1)!}{(r+1)!((n+1)-(r+1))!} = \binom{n+1}{r+1}.\end{align*}$$ 
Therefore,
$$\begin{align*}
(n+1)\sum_{r=0}^{n}(-1)^r\frac{\binom{n}{r}}{r+1} &= \sum_{r=0}^{n}(-1)^r\frac{(n+1)\binom{n}{r}}{r+1}\\
&= \sum_{r=0}^{n}(-1)^r\binom{n+1}{r+1}\\
&= -\sum_{r=0}^n(-1)^{r+1}\binom{n+1}{r+1}\\ 
&= -\sum_{s=1}^{n+1}(-1)^s\binom{n+1}{s}\qquad(\text{setting }s=r+1)\\
&= \left(-\sum_{s=0}^{n+1}(-1)^s\binom{n+1}{s}\right) + (-1)^0\binom{n+1}{0}\\
&= -(1-1)^{n+1} + 1\\
&= 1.
\end{align*}$$
Dividing through by $n+1$ gives the desired result.
Once you realize that 
$$\frac{(n+1)\binom{n}{r}}{r+1} = \binom{n+1}{r+1},$$ it should be obvious that you are dealing with a binomial expansion of some $(n+1)$st power. Then it's just a matter of figuring out which power, and whether any terms are missing. 
