Consider the canonical definition of the gamma function on $\mathbb{C}$:
$$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}dx$$
Problem: How can one show that $\Gamma$ has poles for $z=0, -1, \ldots$?
One proof I've seen uses various facts about the gamma function to show that:
$$\Gamma(z) = {\Gamma(z + n + 1) \over z(z+1) \ldots (z +n)}$$
This clearly shows the result in question; however, this latter representation of the gamma function is "far away" from the former, canonical representation.
Question: Is there a way to show that $\Gamma$ has poles by sticking close to its canonical representation (i.e., by using facts about integrals and so forth)?
The integral representation $$ \Gamma(z)=\int_{0}^{+\infty} x^{z-1} e^{-x}\,dx \tag{1}$$ holds for any $z:\text{Re}(z)>0$. Over such region the functional relation $\Gamma(z+1)=z\,\Gamma(z)$ is a straightforward consequence of integration by parts and Euler's product $$ \Gamma(s)=\lim_{n\to +\infty}\frac{n! n^s}{s(s+1)\cdots(s+n)}\tag{2}$$ is a consequence of the dominated convergence theorem (see my notes, pages 70+). By $(2)$ the $\Gamma$ function is non-vanishing over $\text{Re}(z)>0$ and by $(1)$ it is holomorphic. The functional relation provides an analytic continuation to the complex plane: for instance, since $\Gamma(z)=1+O(z)$ in a neighbourhood of $z=1$, the Laurent series of $\Gamma(z)$ centered at the origin is given by $\frac{1}{z}+O(1)$, since $\Gamma(z)=\frac{\Gamma(z+1)}{z}$. In particular the origin is a simple pole with residue $1$ for the $\Gamma$ function. The same principle leads to the fact that the singularities of the $\Gamma$ function are located at $\{0,-1,-2,-3,\ldots\}$ and these points are simple poles with residues $\{\frac{1}{0!},-\frac{1}{1!},\frac{1}{2!},-\frac{1}{3!},\ldots\}$.