Complex gamma function Let $z$ be a complex number with positive real part. By induction
on $n$, show that
$$
\int_{t=0}^1{t^{z-1}(t-1)^n}dt = \frac{n!}{z(z+1)...(z+n)}.
$$ 
Substitute $t = \frac{u}{n}$ and let $n → ∞$ to obtain
$$
\Gamma \left( z \right) = \lim_{n\to\infty} \frac{n!n^{z-1}}{z(z+1)...(z+n-1)}.
$$ 
I'm having a hard time solving this problem and quite honestly, I'm really rusty on the foundations. I set $n=0$ and got
$$
\int_{[0,1]}t^{z-1}dt=\lim_{a \rightarrow 0} \int_{[a,1]} t^{z-1}\,dt,
$$
and I know I'm supposed to look at 
$$
\frac{d}{dt}(t^z)=zt^{z-1}
$$ 
integrate both sides, and let $a$ go to $0$, but I'm not sure how to go about this.
I know it's silly to not be able to do something so simple on a problem such as this, but I'm just drawing a blank for some reason. I'd be really grateful if someone could help me out.
 A: Well, we start with the easy bit
$$\int_0^1 t^{z-1}(t-1)^n\,dt$$
with $n=0$ we get
$$\int_0^1t^z{dt\over t}$$
set $u=\log t$ so that $du={dt\over t}$ and we get
$$\int_{-\infty}^0e^{uz}\,du={1\over z}e^{uz}\bigg|_{-\infty}^0={1\over z}$$
so this holds for $n=0$. Assume we have tested up through some $n=k-1$, then we seek to compute
$$\int_0^1 t^{z-1}(t-1)^k\,dt$$
Integrating by parts we have
$${1\over z}t^z(t-1)^k\bigg|_0^1+k\int_0^1t^z(t-1)^{k-1}\,dt$$
The first term is just $0$, so we are left with the second which we can compute by noting it is what we already know by induction, only with $z$ replaced by $z+1$, i.e.
$${k\over z}\cdot{(k-1)!\over (z+1)(z+2)\ldots (z+k)}={k!\over z(z+1)\ldots (z+k)}$$
completing the induction.
Now we do the substitution
$$\int_0^1\left({u\over n}\right)^{z-1}\left({u\over n}-1\right)^n\,dt$$
This gives
$${1\over n^{z}}\cdot\int_0^n u^{z-1}\left({u\over n}-1\right)^n\,du={n!\over z(z+1)\ldots (z+n)}$$
Note the $n^{-z}$ term, the $n^{1-z}$ comes from the $\left({u\over n}\right)^{z-1}$ the other $n^{-1}$ comes from $dt ={du\over n}$.
Now we note that as $n\to\infty$ the integral becomes
$$\int_0^\infty u^{z-1}e^{-u}\,du$$
But then if we just move over the $n^{1-z}$ term we get
$$\int_0^\infty u^{z-1}e^{-u}\,du =\lim_{n\to\infty} {n^{z-1}n!\over z(z+1)\ldots (z+n-1)}$$
Here we have only a $n^{z-1}$ because we use that $\lim_{n\to\infty} {n\over (z+n)}=1$ to cancel one $n$ with the last $(z+n)$ factor in the denominator.
as desired.
