[Solved]: Can I simplify log(n+1) before showing that it is in O(log n)?

Simplifying the left-hand side is a good strategy. You aren’t going to find a simpler expression that’s equal to $log(n+1)$, though. However, since you’re only interested in proving that $log(n+1)$ is “not too large”, it can be a good strategy to find an intermediate step that’s larger than $log(n+1)$ but still not too large. That is, try to find a function $f$ such that $log(n+1) le f(n)$ (for sufficiently large $n$), and $f(n) in O(log n)$. (Easy exercise: prove that this is true — if you’ve found a such a function $f$ then $log(n+1) in O(log n)$.) So you can try to fill in the statement $log(n+1) le ldots$ with something simpler on the right-hand side. “Simpler” meaning that in the end it’ll be simpler to prove that this is in $O(log n)$. So keep the definition of $O$ in mind: you’ll need to prove that $f(n) le C log n$ for some $C gt 0$. This suggests trying to find $f$ of the form $C log(n)$. Hint:

$x + 1 le x + x$

Hint #2:

You can say that again.

Of course this approach works for other functions and other bounds. Exercise (open-ended): can you generalize this to $Omega$? To $Theta$? To $o$?