Analytic continuation of Gamma function

I have tried proving the analytic continuation of the gamma function.

I am using the notation, \begin{equation} \Gamma(n) = (n-1)! \end{equation} and \begin{equation} (1) \ \ \Gamma(s) = \int_{0}^{\infty} x^{s-1} e^{-x} dx \end{equation}

I have that the recursion, \begin{equation} (2) \ \ \Gamma(s) = \frac{\Gamma(s+1)}{s} \end{equation}

The definition above says that $\Gamma(s)$ converges for $Re(s) > 0$. I have read on some sources that I can get the strip by strip analytic continuation by first looking at $Re(s)$ in $(0,-1)$ and so on but we will have poles at the negative integers.

I am a little confused. Can I say by (2) that the Gamma function is analytic at all Complex values $s=a+bi$ apart from when s is a negative integer? And then use the step by step technique for looking at negative values -$a+0i$ ?

Any help on the matter would be appreciated.

The integral form is a special case. The definition of the Gamma function occurs as an infinite product (Weierstrass form). We can then express $1/\Gamma(z)$ as an entire function:
$$\frac1{\Gamma(z)} = z e^{\gamma z - G(z)}$$
where $\gamma= \lim_{n \to \infty} \sum_{k=1}^{\infty} \frac1{k} - \log{n}$ is the Euler-Mascheroni constant and
$$G(z) = \sum_{k=2}^{\infty} \frac{(-1)^k \zeta(k)}{k} z^k$$
where $\zeta(k)$ is the Riemann zeta function. The zeroes of the RHS are the poles of $\Gamma(z)$, at the negative integers.