Did I misuse the inductive hypothesis? Note $\log$ is $\log_2$ in this problem.
I am working on the proof of a lemma that my professor used in class.  He said we could verify that it works if we want, but I think I used the inductive hypothesis in such a way that I cannot continue my proof.  Either that or I am missing a simple algebraic manipulation.
The use of my inductive hypothesis is based on the fact that I am dealing with inequalities, so $lg(n) \leq lg(n+1)$ for all $n$.  Bellow is my work so far.
$\sum_{k=2}^{n-1}k\log k \leq \frac{1}{2}n^2 \log n - \frac{1}{8}n^2$
Base step: $2 \leq \frac{9}{2} \log 3 - \frac{9}{8}$
$2 \leq \frac{9}{2} \log 2 - \frac{9}{8} = \frac{27}{8} \leq \frac{9}{2} \log 3 - \frac{9}{8}$
Assume: $\sum_{k=2}^{n-1}k\log k \leq \frac{1}{2}n^2 \log n - \frac{1}{8}n^2$
Show: $\sum_{k=2}^{n}k\log k \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\sum_{k=2}^{n-1}k\log k + n \log n \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\frac{1}{2}n^2 \log n - \frac{1}{8}n^2 + n \log n \leq \frac{1}{2}(n+1)^2 \log (n+1) - \frac{1}{8}(n+1)^2$
$\require{cancel} \cancel{\frac{1}{2}n^2 \log n} - \cancel{\frac{1}{8}n^2} + n \log n \leq \cancel{\frac{1}{2}n^2 \log (n+1)} + n \log (n+1) + \frac{1}{2} \log (n+1) - \cancel{\frac{1}{8}n^2} - \frac{1}{4}n - \frac{1}{8}$
$n \log n \leq n \log (n+1) + \frac{1}{2} \log (n+1) - \frac{1}{4}n - \frac{1}{8}$
At this point I am not sure how to proceed.

If you draw some simple diagrams, given some $f(x) > 0$ and $f'(x) > 0,$ you find
$$ \int_{a-1}^b f(x) dx < \sum_{k=a}^b f(k) <  \int_{a}^{b+1} f(x) dx  $$

You are using $a=2$ and $b=n-1$
This is especially helpful when we know a closed form indefinite integral of $f$
In particular, using logarithm base $e \approx 2.718,$ we have
$$ \int x \log x \; \; dx = \left( \frac{2 x^2 \log x - x^2}{4} \right) + \mbox{constant} $$
NOT SURE: after your edit, I cannot tell which logarithm you are using.
I would say that the constant $2$ matters here, you are using
$$ \frac{\log x}{\log 2} $$
and need to be a little careful about that:
$$ \int \frac{x \log x}{\log 2} \; \; dx = \left( \frac{2 x^2 \log x - x^2}{4 \log 2} \right) + \mbox{constant} $$
