Looking for Guidance: Behavior of $x!^{\frac{1}{x}}$... On my own time, I've been trying to learn as much as I can about the upper levels of mathematics.  I recently came across the Gamma function: 
$$\Gamma(n) = (n-1)! = \int_{0}^{\infty}(t^{x-1}e^{-x})dt = \int_{0}^{1}(-\ln(t))^{x-1}dt$$
Therefore, obviously $x! = \int_{0}^{1}(-\ln(t))^{x-1}dt$ (this can also be verified by graphing both functions).  This reminded me of something I thought about a long time ago: $f(x)=(x!)^{\frac{1}{x}}$.  Now that I understand more about mathematics, I (like many others both in general and on stackoverflow) was able to prove that $(x!)^{\frac{1}{x}}$ diverges to $\infty$.  However, when calculating $\lim_{x\to0}(x!)^{\frac{1}{x}}$, my intial guess that $\lim_{x\to0}(x!)^{\frac{1}{x}}=\gamma=0.57721...$ was proven wrong.  I found that $\lim_{x\to0}(x!)^{\frac{1}{x}}\approx0.5615...$.  This leads me to my first question.  Does this number have any significance?  Might it have any importance other than being the arbitrary number that's the answer to this question?  [ANSWERED BY R_Berger]
Moving on to my second question.  When I graphed this to verify my solution, I was surprised to see that the graph was practically a straight line (I expected more than a negligible curve of some type). Taking the derivative of this function would obviously be shown as $$\frac{d}{dx}\left(\int_0^1\left(-\ln\left(t\right)\right)^{x}dt\right)^{\frac{1}{x}}.$$  I was unable to find the above derivative either by hand or by using a calculator.  So, is there a way to take the above derivative or any other variant of $x!$ (or $\Gamma(x-1)$)?  [NO POINTERS GIVEN YET]
Lastly, most importantly, and my primary reason for asking this question... My ultimate goal is finding $\lim_{x\to\infty}\frac{d}{dx}\left(\int_0^1\left(-\ln\left(t\right)\right)^{x}dt\right)^{\frac{1}{x}}.$  If you have any pointers or small hints on anything I could look into, that would very much be appreciated.  [NO POINTERS GIVEN YET]
If you take the time to read and answer this, I thank you very much in advance.  
 A: For limits of $f(x)^{1/x}$ sometimes one looks at logarithms, so we could look at $$\ln \Gamma(x+1)^{1/x} = \frac{\ln\Gamma(x+1)}{x}.$$
For $x\to0$ this looks like a case for de l'Hospital, for the denominator we get $1$ and for the numerator we have:
$$ \ln \Gamma(x+1)' = \frac{\Gamma(x+1)'}{\Gamma(x+1)}$$ at $x=0$ this evaluates (as you know) to $$\frac{-\gamma}{1}=-\gamma.$$
By exponentiating we recieve our $$e^{-\gamma}=\lim_{x\to0}\exp(\ln\Gamma(x+1)^{1/x})=\lim_{x\to0}\Gamma(x+1)^{1/x}$$

Edit
The case of $x\to\infty$:
Using the Sterling approximation in terms of the $\Gamma$ function
$$ \Gamma(n+1)\approx \sqrt{2\pi n}\big(\frac{n}{e}\big)^n$$ the expression of interest giving the slope is
$$\lim_{n\to\infty}\frac{\Gamma(n+1)^\frac{1}{n}}{n}=\lim_{n\to\infty} \big(\sqrt{2\pi n}\big(\frac{n}{e}\frac{1}{n}\big)^n\big)^\frac{1}{n}=\lim_{n\to\infty} \big(2\pi n\big)^{\frac{1}{n}-\frac{1}{2}} \lim_{n\to\infty} \frac{\big(\frac{n^n}{e^n}\big)^\frac{1}{n}}{n}=$$ 
$$\lim_{n\to\infty} (2\pi)^\frac{1}{n} \lim_{n\to\infty}(n)^{\frac{1}{n}} \lim_{n\to\infty} \frac{1}{e}=e^{-1}.$$
