show that $\sum_{n=1}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k+1}\binom{n}{k}\log(k+1)=1$ $$1=\sum_{n=1}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}(-1)^{k+1}\binom{n}{k}\log(k+1)$$
I tried by several ways but failed -_- 
the expression is another form to e number by take exponential of both side 
 A: I'm pretty sure this formula is actually incorrect: I would expect the value of the sum to be the Euler-Mascheroni constant $\gamma$.
Explanation: write $\log{(k+1)}= \left. -\frac{d}{ds}\right|_{s=0} (1+k)^{-s}$. Then we have the integral representation
$$ \frac{1}{(1+k)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} e^{-(1+k)x} \, dx, $$
and swapping the order of summation and integration, we have
$$\begin{align*}
\sum_{k=0}^{n} &(-1)^{k+1}\binom{n}{k} \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} e^{-(1+k)x} \, dx \\
&= \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} \left( \sum_{k=0}^{n} (-1)^{k+1}\binom{n}{k} e^{-(1+k)x} \right) \, dx \\
&= \frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1} \left( -e^{-x}(1-e^{-x})^n \right) \, dx
\end{align*}$$
Doing this again for the outer sum (and noting the $n=0$ term is zero in the sum we want, so we throw it in for ease of calculation), we have
$$\begin{align*}
\sum_{n=0}^{\infty} & \frac{1}{n+1} \frac{1}{\Gamma(s)} \int_0^{\infty} x^{s-1} \left( -e^{-x}(1-e^{-x})^n \right) \, dx \\
&= -\frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1}e^{-x} \left(\frac{1}{1-e^{-x}}\sum_{n=0}^{\infty} \frac{1}{n+1}(1-e^{-x})^{n+1} \right) \, dx \\
&= -\frac{1}{\Gamma(s)}\int_0^{\infty} x^{s-1}e^{-x} \left(\frac{1}{1-e^{-x}}\left(-\log{(1-(1-e^{-x}))} \right)\right) \, dx \\
&= -\frac{1}{\Gamma(s)}\int_0^{\infty} \frac{x^s}{e^{x}-1} \, dx \\
&= -s\zeta(1+s)
\end{align*}$$
Therefore we just have to find the coefficient of $s$ in the Taylor series of $s\zeta(1+s)$ at $s=0$. This expansion is well-known to be
$$ 1 + \gamma s + O(s^2), $$
so the answer is $\gamma$ as I claimed.
(There is admittedly a worryingly large amount of limit interchange going on here.)
For a few other proofs, see Jonathan Sondow's 2003 paper on arXiv.
A: By Frullani's theorem we have:
$$-\log(k+1)=\int_{0}^{+\infty}(e^{-kx}-1)\frac{dx}{x e^x} $$
hence:
$$ \sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}\log(k+1)=\int_{0}^{+\infty}(1-e^{-x})^n\frac{dx}{x e^x}$$
as well as:
$$ \sum_{n\geq 1}\frac{1}{n+1}\sum_{k=0}^{n}\binom{n}{k}(-1)^{k+1}\log(k+1)=\int_{0}^{+\infty}\frac{1-e^x(1-x)}{e^x-1}\frac{dx}{x e^x}=\color{red}{\gamma}$$
that is the Euler-Mascheroni constant.
