The Euler Gamma function is defined by
$$\begin{align} &\Gamma : \{z\in \mathbb{C} : \Re(z) > 0\} \rightarrow \mathbb{C} \\ &\Gamma(z) = \int_{0}^{\infty} x^{z-1}e^{-x}dx \end{align}$$
My goal is to show that $\Gamma$ is continuous. I will quote the solution from my exercsce sheet below:
Let $z_n \rightarrow z_0, \Re(z_k) > 0 \; \forall k \geq 0$. We can assume that $|z_n| \leq 1$ without loss of generality. By the Dominated Convergence Theorem: $$\lim_{n\rightarrow\infty}\int_{0}^{\infty} x^{z_n-1}e^{-x}dx = \int_{0}^{\infty} x^{z_0-1}e^{-x}dx$$ since $\int_{0}^{\infty}e^{-x}dx$ is integrable.
In this solution the approach seems to be that for any sequence $z_n$ converging to $z_0$, $\lim_{n \rightarrow \infty}\Gamma(z_n) = \Gamma(z_0)$, but I don't understand why it can be assumed that $|z_n|\leq1$.
Secondly, for applying DCT, $g(x) = e^{-x}$ was used. This only works if for the absolute value of the integrand: $|x^{z_n-1}e^{-x}| \leq g(x)$. But why is this true? I tried:
$$|x^{z_n-1}e^{-x}| = e^{z_n\log{x}}x^{-1} e^{-x} = e^{\Re(z_n)\log{x}}x^{-1}e^{-x} = x^{\Re(z_n)-1}e^{-x} ~,$$ where the absolute value vanishes because $x\in(0,\infty)$.
Now why is $x^{\Re(z_n)-1} \leq 1$?