Proof that $ \log_2(n^2+n) = \Omega(\log_2(n)) $

Question

Problem

Proof that $ \log_2(n^2+n) = \Omega(\log_2(n)) $

Attempt to proof

$ \log_2(n^2+n) = \Omega(\log_2(n)) $ hold when $\exists(c,n_0)$ for which (by definition) $$ \log_2(n) \ge c\log_2(n^2+n), \text{ when } n \ge n_0 $$ let $c=0, n_0=1$ $$ \implies \log_2(n) \ge 0, \text{ when } n \ge 1$$ Which holds for $\forall n$.

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Is my proof correct?

Answer 1

$$\log(n^2+n)>\log(n^2)=2\log(n).$$