Derivative of $\Gamma$ at $1$ I've been given two definitions of the Gamma function, the integral defintion:
$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$ (for $Re(z)>0$)
and the product definition (for $1/\Gamma$):
$\frac{1}{\Gamma(z)} = ze^{\gamma z}\Pi_{n=1}^\infty ((1+\frac{z}{n})e^{-\frac{z}{n}})$
where $\gamma$ is Euler's constant
My lecturer has asserted that therefore (presumably from the product definition): 
$\Gamma'(1) = -\gamma$
but I can't see why this is true. Is this something that follows easily from these definitions? If it is I would appreciate some help or a solution. Thanks :)
 A: Start from the product form and take the log of both sides, and then differentiate:
(Sorry, I have to write out the equations in LaTeX format, I don't know MathML well enough)
Take the log:
\begin{equation}
  - \log (\Gamma(z)) = \log(z) + \gamma z
  + \sum_{n=1}^{\infty} \left[ \log \left( 1 + \frac{z}{n} \right) - \frac{z}{n} \right]
\end{equation}
Take the derivative of both sides with respect to z:
\begin{equation}
- \frac{\Gamma^{\prime}(z)}{\Gamma(z)} = \frac{1}{z} + \gamma 
 + \sum_{n=1}^{\infty} \left[ \frac{1/n}{1+z/n} - \frac{1}{n} \right]
\end{equation}
Now insert z = 1 because we only care about the derivative at z = 1, and simplify the fraction with 1+1/n in the denominator (using $\Gamma(1) = \int_0^\infty e^{-t}dt = 1$):
\begin{align}
- \Gamma' (1) = 1 + \gamma + \sum_{n=1}^{\infty} \left[ \frac{1}{n+1} - \frac{1}{n} \right] = 1 + \gamma - \sum_{n=1}^{\infty} \frac{ 1}{n(n+1)} = \gamma. 
\end{align}
which is $\Gamma'(1) = - \gamma$.
A: Here if we use the integral definition we get:
$\Gamma'(z)=\int_0^\infty \frac{\partial}{\partial z}(t^{z-1})e^{-t}dt=\int_0^\infty \ln(t)t^{z-1}e^{-t}dt$
thus $\Gamma'(1)=\int_0^\infty \ln(t)e^{-t}dt=-\gamma$
