Show that this sequence converges to $0$

Let me construct an elementary proof.

Let $\sqrt[2k']{1+a}=1+b$. Then, $b>0$ since $a>0$, and $$ x_n=\frac{n^{k'}}{(1+a)^n}=\frac{n^{k'}}{(1+b)^{2k'n}}=\left(\frac{\sqrt{n}}{(1+b)^{n}}{}\right)^{2k'} $$ But $$ (1+b)^n\ge 1+bn>bn, $$ and thus $$ \frac{1}{(1+b)^n}<\frac{1}{bn}, $$ and finally $$ x_n=\left(\frac{\sqrt{n}}{(1+b)^{n}}{}\right)^{2k'}<\left(\frac{\sqrt{n}}{bn}\right)^{2k'}=b^{-2k'}\cdot\frac{1}{n^{k'}} $$ Now, it suffices to show that the right hand side tends to zero, as $n$ tends to infinity.