How to show $\sum_{n=0}^m \frac{1}{n+1}\binom{m}{n} = \frac{2^m-1}{m+1}$ This is the homework, and it shouldn't be difficult, but I can't find the proper identity that would help me simplify this sum:

$$\sum_{n=0}^m \frac{1}{n+1}\binom{m}{n}$$

Through calculating the results, I can see that the simplified version is:

$$\frac{2^m-1}{m+1}$$

But I don't know how to transform the former into the later. You need not give the complete solution (although, that's welcomed too), but the identities needed for the simplification should suffice.

EDIT:
How I counted: 
$\frac{m!}{(n+1)!(m-n)!}$ repeated $m$ times while $n$ increases from 0 to $m$. You can also see the code here: http://pastebin.com/RJ9jd966
 A: Hint. Consider the function
$$
\sum_{n=0}^m\frac{1}{n+1}{m\choose n} x^{n+1}.\tag{1}
$$
Differentiating it gives you something very familiar.

Since you've already accepted an answer, I'll work out my answer so that you can see another approach. Differentiating the above polynomial gives
$$
\sum_{n=0}^m{m\choose n}x^n
$$
which may be recognized as a special case of Newton's binomial series: it is equal to $(x+1)^m$. The primitive of this polynomial is $p(x)=\frac{1}{m+1}(x+1)^{m+1}+c$, where we have to choose the constant such that $p$ agrees with (1). One way to do this, is by looking at the value at $0$. 
$$
p(0)=\frac{1}{m+1}+c
$$
while the function in (1) gives $0$ at $0$. This means that we must choose $c=-\frac{1}{m+1}$. So we obtain
$$
\sum_{n=0}^m\frac{1}{n+1}{m\choose n} x^{n+1}=\frac{(x+1)^{m+1}}{m+1}-\frac{1}{m+1}
$$
evaluating the latter term at $1$ gives the answer:
$$
\frac{2^{m+1}-1}{m+1}.
$$
A: Hints:


*

*For every $0\leqslant n\leqslant m$, $\displaystyle\frac1{n+1}{m\choose n}=\frac1{m+1}{m+1\choose n+1}$.

*$\displaystyle\sum\limits_{n=0}^m{m+1\choose n+1}=2^{m+1}-1$.


Note: Hence the result is not $\dfrac{2^m-1}{m+1}$ but a slight modification.
