Simplifying $\sum\limits_{k=1}^{n-1} (2k + \log_2(k) - 1)$ I'm trying to simplify the following summation:
$$\sum_{k=1}^{n-1} (2k + \log_2(k) - 1)$$.
I've basically done the following:
$$\sum_{k=1}^{n-1}  (2k + \log_2(k) - 1) \\
=
\sum_{k=1}^{n-1} 2k + \sum_{k=1}^{n-1} \log_2(k) - \sum_{k=1}^{n-1} 1\\
=
\frac{n(n-1)}{2} + \sum_{k=1}^{n-1} \log_2(k) - (n-1)$$
Now I'm trying to do deal with this term $\sum_{k=1}^{n-1} \log_2(k)$, but I'm a bit confused.
My gut tells me I can do the following:
$$\sum_{k=1}^{n-1} \log_2(k)\\
=
\log_2(1) + \log_2(2) + \ldots + \log_2(n-1)\\
= 
\log_2(\prod_{k=1}^{n-1} k)\\
= 
\log_2((n-1)!)$$
Using that $\log_a(b) + \log_a(c) = \log_a(b \cdot c)$.
However, I'm not convinced this is an entirely valid reasoning because I can't find any rules/identities for dealing with $\sum_{k=1}^{n-1} \log_2(k)$.
Is this correct or are there any rules to apply?

(Just so people know this has been answered)
You right 
$$\sum_{k=1}^{n} \log k = \log (n!)$$
If you want to justify it formally, you can try using induction.
