Proof of convergence of integral representation of the Gamma Function The Gamma Function defined by :    $\Gamma(n) = \displaystyle\int_0^\infty e^{-x} x^{n-1} dx$    This can be written as:   $\displaystyle\Gamma(n) = \int_0^1 e^{-x} x^{n-1} dx + \int_1^\infty e^{-x} x^{n-1} dx$    Now the convergence of the $1$st first integral is clear to me but I am not sure about the $2$nd integral. Our professor did the proof in the following way :  $\displaystyle e^x = 1+x+\frac{x^2}{2!}+\ldots+\frac{x^{n+1}}{(n+1)!}+\ldots  > \frac{x^{n+1}}{(n+1)!}\\$ $\implies e^{-x} x^{n-1} < \frac{{(n+1)!}{x^{n-1}}}{x^{n+1}} = \frac{(n+1)!}{x^2}\\$ Integrating both sides from 1 to infinity :$\displaystyle\int_1^\infty e^{-x} x^{n-1}dx < \small(n+1)! \int_1^\infty \frac{dx}{x^2}\\$ Now since $\displaystyle\int_1^\infty \frac{dx}{x^2}$ is convergent ( I have that proof ) So by Comparison test $\displaystyle\int_1^\infty e^{-x} x^{n-1}dx$ is also convergent. I think the proof is wrong because firstly while expanding the series of $e^x$ how can she write $\frac{x^{n+1}}{(n+1)!}$ , this term using $n$ because till then I have no idea about the value of $n$ and writing this term means that $n \geq 0$ and here on the contrary we need to find the restriction over $n$ for which this integral converges. If this step is correct then the rest of the proof is pretty simple to unterstand. Can anyone tell me whether the proof is correct or if it is wrong, in which part it is and how to do it correctly ?
$Edits: \\ \\$
Most of the comments and answers mentioned that my professor must have assumed $n$ to be $>0$. But that's not the case. Instead she intends to find the interval of $n$ for which $\Gamma$ - Function converges. From the first part i.e. the integral from 0 to 1 she derived the condition $n > 0$. But that has no connection with the $2$nd part i.e. the integral from $1$ to $\infty$. She didn't assume any restriction on $n$ . That's why I asked this question here. I request everyone to please add the proof of the 2nd part without assuming any restriction on $n\\$.  Thanks for the help.
 A: For $\text{Re}(z)>0$, we can represent the Gamma Function by the integral
$$\Gamma(z)=\int_0^\infty e^{-x} x^{z-1}\,dx$$

For any integer $n$ and $x\ge0$, we have the bound
$$e^x\ge \frac{x^{n+1}}{(n+1)!}$$

Now, we fix the number $z$ and we choose an integer $n\ge \max(1,\text{Re}(z)+1)$.  Then, we can assert
$$\begin{align}
\left|\int_1^\infty e^{-x}x^{z-1}\,dx\right|&\le \int_1^\infty e^{-x}x^{\text{Re}(z)-1}{x}\,dx\\\\
&\le \int_1^\infty \frac{x^{\text{Re}(z)-1}}{\frac{x^{n+1}}{(n+1)!}}\,dx\\\\
&=(n+1)!\int_1^\infty \frac{1}{x^{1+(n-\text{Re}(z))}}\,dx\\\\
&\le (n+1)!\int_1^\infty \frac1{x^2}\,dx
\end{align}$$
And the integral converges for this fixed $z$.

Of course, this fixed value of $z$ was arbitrary.  So, for any other $z$, we can choose an integer $n\ge \max(1,\text{Re}(z)+1)$ and arrive at the same conclusion that $\int_1^\infty e^{-x}x^{z-1}\,dx$ converges.
A: The asymptotic behavior of the integrand is
$$\sim x^{n-1}$$ near $0$, which integrates as $$\sim\frac{x^n}n,$$ ($\log x$ for $n=0$) and
$$\lesssim e^{-x}\max(x^{n-1},1)$$ near $\infty$, which integrates like $$\lesssim p_n(x)e^{-x}$$ for $n>0$ where $p_n$ is a polynomial of degree $\max(n,0)$.
Hence, the integral in $(1,\infty)$ causes no problem, while that in $(0,1)$ only works for $n>0$.
