According to googology.wikia, we have the following:
$$5^{4^{3^{2^{1}}}}=620606987866087447074832055728467\ldots$$
$$6^{5^{4^{3^{2^{1}}}}}=110356022591769663217914533447534\ldots$$
How are the first digits of these numbers calculated? The last digits are trivial with basic modular arithmetic techniques, but I have no idea how the first digits are found.
On
Actually, it appears this is not very hard to answer. Notice that:
$$5^{4^{3^{2^1}}}=5^{262144}=10^a$$
We solve for $a$ by taking logarithms:
$$a=262144\log_{10}(5)=183230.7928166\ldots$$
Thus, there are $183231$ digits and the first few digits are
$$10^{0.7928166\ldots}=6.206069002\ldots$$
Though you'll have to excuse the accuracy of my hand calculators. Likewise, with stronger computers, the first few digits of $6^{5^{4^{3^{2^1}}}}$ is doable by the same method.
$$4^{3^2} = 262144$$
$$ \log_{10}(5^{4^{3^2}}) = 4^{3^2} \log_{10}(5) \approx 183230.792816661313561889631182$$ so $$ \eqalign{5^{4^{3^2}} &\approx 10^{183230.792816661313561889631182}\cr &\approx 6.20606987866087447074832995296 \times 10^{183230}} $$ In a good CAS, we should be able to get $5^{4^{3^2}}$ as an exact integer with $183231$ digits: call it $N$. Then $\log_{10}(6^N) = N \log_{10}(6)$ can be computed in floating point with a few more than $183231$ digits: if the result is $a.b$, and $c = 10^{0.b}$, then $$ 6^N \approx c \times 10^{a}$$ Using Maple with 183300 digits, I get $$ 0.b = 0.042796039418829008454171873916897558787404373143752795118675574544262$$ and then indeed $$ 10^{0.b} \approx 1.103560225917696632179145334475344911419051187213486921746877175936 $$