Solving the recurrence T(n) = 4T(n/4) + n log n with the iterative method I'm trying to solve

$$ T(n) = 4T(n/4) + n \log_{10}n.$$

I'm having trouble with Iteration Method near the end.
As far as I went, I obtained the General Formula as:

$$4^kT(n/4^k)+n\log n+\sum (n/4^k)\log(n/4^k)$$

And trying to get the moment when it finishes iterating:

$$(n/4^k)=1$$
$$n=4^k$$
$$\log_4n=k$$

And here I get stuck. I know I have to substitute $k$ with $\log_4n$ but after that I'm lost. Can I get a bit of help with explanation of every step? 

Here are some more details:

Swapped $\log_4n$ on all $k$:
$$4^{\log_4n} T(n/4^{\log_4n})+n\log_2n+\sum_{n=0}^{\log_4n}n\log n$$

From logarithm rules of $a^{log_an} = n$, it ends like this:
$$(n)(1)+n\log_2n+\sum_{n=0}^{\log_4n}n\log n$$

I'm not sure how to express the sum in $n$, but as you can see already, it is $O(n\log_2n)$, and with the Master Theorem, you obtain the same result $O(n\log n)$.
Your "General Formula" is incorrect.  The third term (with the sum) is incorrect.  It should be $n$ rather than $n/4^k$.

Make the suitable correction through the rest of your answer, and use the fact that

$$\sum_{k=0}^{\log_4 n} n \log_{10}(n/4^k) \le \sum_{k=0}^{\log_4 n} n \log_{10} n$$

and you should be able to get an upper bound from there.  (The final sum is easy to evaluate as every term is the same.)