show: $$\sum_{i=1}^n \log_{2}\,i = O(n\log n)$$
Proof by induction:
$$\sum_{i=1}^n \log\,i \le n\log n$$
$$\text{Test for n=1: }\sum_{i=1}^1 \log_{2}\,i \le 1\log 1$$ $$0 \le 0\text{ true for }n=1$$
Assume true for $n=k$: $$\sum_{i=1}^k \log\,i \le k\log k$$
Prove true for $n =k+1$: $$\sum_{i=1}^{k+1} \log\,i \le (k+1)\log(k+1)$$
By assumption: $$\log 1 + \log 2 + \ldots + \log k \le k\log k$$
then add the extra $k+1$ term: $$\log 1 + \log 2 + \ldots + \log k + \log(k+1) \le k\log k + \log(k+1)$$
$$\sum_{i=1}^{k+1} \log i \le k\log k + \log(k+1)$$
Not sure how to show: $k\log k + \log(k+1) \lt (k+1)\log(k+1)$ (this shows by the transitivity of $\lt$ that the statement is also true for $n =k+1$)
therefore: $$\sum_{i=1}^{k+1} \log\,i \le (k+1)\log(k+1)$$
Extra: In original problem I see that they had $\log$ to the base $2$ on the left hand side but not on the right hand side. I chose to drop the base but not sure if this was a correct decision.
Since the question as-stated does not say to do induction, I'll do an easier way, since it seems likely you only are doing it that way because you think it best.
$1$) Note $\log_2 x={1\over \log 2}\log x$ by the change of base formula.
$2$) Note that $\log x$ is an increasing function, hence $0<x\le y\iff \log x\le \log y$ and so
$$\sum_{i=1}^n\log_2 i={1\over\log 2}\sum_{i=1}^n\log i\le {1\over \log 2}\sum_{i=1}^n\log n= Cn\log n$$
which completes the proof.
Insofar as your original one goes, if you want to show
$$k\log k+\log(k+1)\le (k+1)\log(k+1)$$
it's not bad, you just use the monotonicity property I noted in ($2$) to see that $k\log k< k\log(k+1)$ and you're done.