Show that for any constants $k$ and $j$, where $j > 0$, $(n+k)^j = \Theta(n^j)$

Question

Does this solve the question: There exist $c, n_0$, such that $(n+k)^j = n^j$ for all $n>=n_0$

I tired the above but couldn't find an answer..

Answer 1

$(n+k)^j = \theta(n^j)$ means

$0< \lim \inf \frac{(n+k)^j}{n^j} \le \lim \sup \frac{(n+k)^j}{n^j}< \infty$.

But this is clear since $\frac{(n+k)^j}{n^j} \to 1$ as $ n \to \infty$.