Approximate $\log_{10}$ values without a calculator I've got this problem:

$1,000,000^{{1,000,000}^{1,000,000}} < n^{n^{n^n}}$
What is the first positive integer value of n for which this inequality holds?

I managed to reduce it to this:

$6+\log_{10}(6) < n\log_{10}(n)$

by using $\log_{10}$ three times (and discarding some of the insignificant values – I can explain why these are insignificant at the end).
The only problem is I don't know how to approximate this (using an upper bound on the left and lower on the right as an upper bound for n, and then the reverse for a lower bound of n) accurately enough.
Does anyone have any ideas?
Also, if anyone has another way of doing this problem without the method I used, avoiding my issue altogether, that would be helpful.

Here is a solution with limited hand multiplications/additions. Starting from your condition that $6+\log_{10} 6< n\log_{10} n$, notice that
$$
10<6^2<100
\rightarrow 1< 2\log_{10} 6 < 2
\rightarrow 1/2 < \log_{10} 6 < 1
\longrightarrow \fbox{$6.5 < 6+\log_{10} 6 < 7$}.
$$
Also, $6^6=46,656$ (by hand!) therefore
$$
10^4<6^6<10^5
\rightarrow 4 < 6\log_{10} 6 < 5
$$
therefore $n=6$ is too small. On the other hand
$$
\text{with } n=10\quad 10\log_{10} 10 = 10 > 7
$$
therefore $n=10$ is large enough. Between 6 (too small) and 10 (large enough), try 8 :
\begin{align}
& 8^8 = (2^3)^8=2^{24}=(2^{10})^2\times 2^4 > 1000^2\times 16\\
\longrightarrow\quad 
&
8\log_{10} 8 > 2\log_{10} 1000 + \log_{10} 16 > 6+1=7
\end{align}
therefore 8 is also enough. The only remaining possibility is 7:
\begin{align}
& 7^2=49<50 
\rightarrow
7^6<50^3=125,000
\rightarrow
7^7<7\times 125,000=875,000 < 10^6\\
\longrightarrow\quad &
7\log_{10} 7 < 6 
\end{align}
and thus $n=7$ is not enough. The answer is 8.
PS. Using Knuth's up-arrow notation, you get $7\uparrow\uparrow4<(10^6)\uparrow\uparrow3<8\uparrow\uparrow4$.
