Taking the limit outside the integral I'm trying to work through Gauss' derivation of the Gamma function from: 
$$
\Gamma : \mathbb{R^+}\mapsto\mathbb{R} \;\;,\;\;\Gamma(n) = \int_0^\infty y^{n-1}e^{-y}dy\,.
$$
I know the limit expression for $e^{-y}$ is 
$$
e^{-y} = \lim_{m\rightarrow\infty}\bigg(1-\frac{y}{m}\bigg)^m.
$$
If I then have 
$$
\Gamma : \mathbb{R^+}\mapsto\mathbb{R} \;\;,\;\;\Gamma(n) = \int_0^\infty y^{n-1}\lim_{m\rightarrow\infty}\bigg(1-\frac{y}{m}\bigg)^mdy\,.
$$
Am I then able to take out the limit from the integral? What are the conditions/justification of being able to do so? Is it possible to extend this to a complex domain? Would it be possible if the domain was the reals without negative integers?
Thanks.
 A: The dominated convergence theorem can be used here to justify switching the limit and the integral but some care with the integration limits is required.  
Note that the integral of $y^{n-1}(1 - y/m)^m$ over $[0,\infty)$ is divergent and the result you are looking for is
$$\lim_{m \to \infty}\int_0^m y^{n-1} \left(1 - \frac{y}{m} \right)^m \, dy = \int_0^\infty y^{n-1} e^{-y} \, dy. $$
We can justify this first by putting the upper limit on the left-hand side  to $\infty$ after multiplying the integrand by the indicator function $\chi_{[0,m]}(y)$ which is defined as
$$\chi_{[0,m]}(y) = \begin{cases}1, \,\,\,0 \leqslant y \leqslant m\\ 0, \,\,\,y > m  \end{cases}$$
This gives us
$$\begin{align}\lim_{m \to \infty}\int_0^m y^{n-1} \left(1 - \frac{y}{m} \right)^m \, dy &= \lim_{m \to \infty}\int_0^\infty y^{n-1} \left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \, dy \\ &= \int_0^\infty \lim_{m \to \infty}y^{n-1}  \left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \, dy \\ &= \int_0^\infty y^{n-1} e^{-y} \, dy \end{align}.$$
The dominated  convergence theorem has been applied to justify the interchange for an integrand that is both integrable over $[0,\infty)$ and dominated by the integrable function to which it converges. Note that as $m \to \infty$ we have
$$\left(1 - \frac{y}{m} \right)^m \chi_{[0,m]}(y) \uparrow e^{-y},$$
since the LHS increases monotonically with respect to $m$.  An application of Bernoulli's inequality verifies this.
Since $m + 1 > m$ we have
$$\left(1 - \frac{y}{m+1} \right)^{(m+1)/m} \geqslant  1 - \frac{y}{m+1}\frac{m+1}{m}  =  1 - \frac{y}{m} \\  \implies \left(1 - \frac{y}{m+1} \right)^{m+1} \geqslant \left(1 - \frac{y}{m} \right)^{m}.$$
The integral representation of $\Gamma(z)$ where $z \in \mathbb{C}$ is convergent only for $\Re(z) > 0.$
