In order to prove the above result I proceeded as follows:
We know that:
$\Gamma(n)=(n-1)\Gamma(n-1)$
Using this fact, we have:
$\Gamma(n+k)=(n+k-1)\Gamma(n+k-1)\\=(n+k-1)(n+k-2)\Gamma(n+k-2)\\
=...=(n+k-1)(n+k-2)...(n+k-k)\Gamma(n+k-k)\\=n(n+1)...(n+k-1)\Gamma(n)$
So, now we have: $$\lim\limits_{n\rightarrow\infty}\frac{\Gamma(n+k)}{\Gamma(n)}= \lim\limits_{n\rightarrow\infty}\frac{n(n+1)...(n+k-1)\Gamma(n)}{\Gamma(n)}$$
Now, if $x$ is a variable and $c$ is a constant, then
$x+c\approx x$ (for very large values of $x$) (I am a little doubtful about this)
Using this fact, can we say that:
$\lim\limits_{n\rightarrow\infty}\frac{\Gamma(n+k)}{\Gamma(n)}=\lim\limits_{n\rightarrow\infty}n(n+1)...(n+k-1)\\\sim n.n...n\ (k\ times)\\=n^k$
Thus, we have: $$\frac{\Gamma(n+k)}{\Gamma(n)}\sim n^k$$
Is this method valid? Is there a more "mathematically rigorous" way to prove this?
Thanks in advance!
Instead of trying to prove an approximation $$ \frac{\Gamma(n+k)}{\Gamma(n)}\approx n^k $$ prove the more precise asymptotics: $$ \frac{\Gamma(n+k)}{\Gamma(n)}\sim n^k $$ where by definition $$ f(x) \sim g(x) \quad\text{iff}\quad \lim_{x\to\infty} \frac{f(x)}{g(x)} = 1 $$ With this, your proof works fine, because $$ n(n+1)\cdots(n+k-1) \sim n^k $$
Wikipedia says that $$\lim_{n\to\infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^{\alpha}} = 1 \quad\text{for all } \alpha\in\mathbf{R},$$ which means that $$ \frac{\Gamma(n+\alpha)}{\Gamma(n)}\sim n^\alpha $$ but the proof is likely to be different from the one above.