Asymptotic approximation regarding the Gamma function $\Gamma$.
The most usual derivation of this would involve the Stirling-Laplace asymptotic for $\Gamma(s)$. I'm mildly surprised that this wasn't explicitly worked out in Wiki, or some other easily accessible places on-line.
In fact, a much simpler approach obtains (a stronger version of) this asymptotic via "Watson's Lemma", which is itself easy to completely prove from simple things, going back over 100 years. In various places in the literature, the lemma is in fact called "the oft-reproven Watson's lemma". :)
The case you mention is a simple corollary of the very first example I wrote out in some notes on asymptotic expansions: http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/02d_asymptotics_of_integrals.pdf
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Stirling Asymptotic: $\ds{N! \sim \root{2\pi}\, N^{N + 1/2}\expo{-N}}$ as $\ds{\verts{N} \to \infty}$.
\begin{align} \left.\lim_{n \to \infty}{\Gamma\pars{n + \alpha} \over \Gamma\pars{n}n^{\alpha}}\,\right\vert_{\ \alpha\ \in\ \mathbb{C}} & = \lim_{n \to \infty}{\pars{n + \alpha - 1}! \over \pars{n - 1}!\, n^{\alpha}} \\[5mm] & = \lim_{n \to \infty}{\root{2\pi}\pars{n + \alpha - 1}^{n + \alpha - 1/2}\expo{-\pars{n + \alpha - 1}} \over \bracks{\root{2\pi}\pars{n - 1}^{n - 1/2}\expo{-\pars{n - 1}}}\, n^{\alpha}} \\[5mm] & = \lim_{n \to \infty}{n^{n + \alpha - 1/2}\, \bracks{1 + \pars{\alpha - 1}/n}^{n + \alpha - 1/2}\,\expo{-\alpha} \over \bracks{n^{n - 1/2}\pars{1 - 1/n}^{n - 1/2}}\, n^{\alpha}} \\[5mm] & = \expo{-\alpha}\lim_{n \to \infty} {\bracks{1 + \pars{\alpha - 1}/n}^{n} \over \pars{1 - 1/n}^{n}} \\[5mm] & = \expo{-\alpha}\,{\expo{\alpha - 1} \over \expo{-1}} = \bbx{1} \end{align}
Real $\boldsymbol{\alpha}$
The log-convexity of the Gamma function is shown in this answer.
Suppose that $0\le\alpha\le k\in\mathbb{Z}$, then using the recurrence relation for $\Gamma$, $$ \begin{align} \Gamma(n+\alpha) &\le\Gamma(n)^{1-\alpha/k}\,\Gamma(n+k)^{\alpha/k}\\ &\le\Gamma(n)^{1-\alpha/k}\left(\Gamma(n)\,(n+k-1)^k\right)^{\alpha/k}\\ &=\Gamma(n)\,(n+k-1)^\alpha\tag1 \end{align} $$ and $$ \begin{align} \Gamma(n) &\le\Gamma(n+\alpha-k)^{\alpha/k}\Gamma(n+\alpha)^{1-\alpha/k}\\[6pt] &\le\left(\frac{\Gamma(n+\alpha)}{(n+\alpha-k)^k}\right)^{\alpha/k}\Gamma(n+\alpha)^{1-\alpha/k}\\ &=\frac{\Gamma(n+\alpha)}{(n+\alpha-k)^\alpha}\tag2 \end{align} $$ Then we have $$ \left(\frac{n+\alpha-k}{n}\right)^\alpha \le\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha} \le\left(\frac{n+k-1}{n}\right)^\alpha\tag3 $$ and by the Squeeze Theorem, for $\alpha\ge0$, $$ \lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}=1\tag4 $$ Furthermore, $$ \begin{align} \Gamma(n) &\le\Gamma(n-\alpha)^{1-\alpha/k}\Gamma(n-\alpha+k)^{\alpha/k}\\ &\le\Gamma(n-\alpha)^{1-\alpha/k}\left(\Gamma(n-\alpha)(n-\alpha+k-1)^k\right)^{\alpha/k}\\ &=\Gamma(n-\alpha)(n-\alpha+k-1)^\alpha\tag5 \end{align} $$ and $$ \begin{align} \Gamma(n-\alpha) &\le\Gamma(n)^{1-\alpha/k}\Gamma(n-k)^{\alpha/k}\\ &\le\Gamma(n)^{1-\alpha/k}\left(\frac{\Gamma(n)}{(n-k)^k}\right)^{\alpha/k}\\ &=\frac{\Gamma(n)}{(n-k)^\alpha}\tag6 \end{align} $$ Therefore, $$ \left(\frac{n-\alpha+k-1}{n}\right)^{-\alpha} \le\frac{\Gamma(n-\alpha)}{\Gamma(n)\,n^{-\alpha}} \le\left(\frac{n-k}{n}\right)^{-\alpha}\tag7 $$ and by the Squeeze Theorem, for $\alpha\ge0$, $$ \lim_{n\to\infty}\frac{\Gamma(n-\alpha)}{\Gamma(n)\,n^{-\alpha}}=1\tag8 $$
Complex $\boldsymbol{\alpha}$
Unfortunately, I have not found a way to make the log-convexity argument that works for $\alpha\in\mathbb{R}$ work for $\alpha\in \mathbb{C}$. About the best I can see, is to use Stirling's Approximation. $$ \Gamma(n)\sim\sqrt{\frac{2\pi}n}\frac{n^n}{e^n}\tag9 $$ Applying $(9)$ to $\Gamma(n+\alpha)$ and $\Gamma(n)$, we get $$ \frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}\sim\sqrt{\frac{n}{n+\alpha}}\frac{\left(1+\frac\alpha{n}\right)^{n+\alpha}}{e^\alpha}\tag{10} $$ which yields $$ \lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}=1\tag{11} $$