Approximation of (n^n)^n

Question

To be specific, what is the best way to calculate the first 10 digits decimal approximation of $$ \large \left(123456789^{123456789}\right)^{123456789}$$?

Even WolframAlpha gives the result in a power of 10 representation as $$ \large 10^{10^{10^{1.232768993649683}}}$$ Is there any other ways to approximate the acceptable first 10 digits in decimal?

Answer 1

Let $m = (n^n)^n = n^{n^2}$. Then, taking logarithms in base 10, $\log{m} = n^2 \log{n}$. Therefore we can write

$$ m = s \cdot 10^p$$ where $s \in (1,10)$ is given by exponentiating the fractional part of $n^2 \log{n}$, and $p$ is the integer part of $n^2 \log{n}$. The leading ten digits of $m$ are encoded in $s$.

Answer 2

With $n=123456789$ the first 10 decimal digits of $m=(n^n)^n=n^{n^2}$ can be computed as follows (using Pari/GP with 60 decimal digit precision):

$$a = \log_{10}m = n^2 \log_{10}n$$ $$a=123327462732871491.130863690559566545920203026360790125577391$$ (The integer part of $a$ explains the $1.23327\times 10^{17}$ decimal digits from your Wolfram link). The leading digits of $m$ come from the fractional part $f$ of $a$ $$f \approx0.130863690559566545920203026360790125577391$$ and are computed as $10^f$

$$m=10^f\times 10 ^{123327462732871491}$$ $$m\approx 1.351648262765413474237868427278 \times 10^{123327462732871491}.$$

Thus the leading 10 digits of $(123456789^{123456789})^{123456789}$ are $1351648262.$


Here an example with smaller $n=7$ where you can see the complete number $m=(n^n)^n$ $$m=256923577521058878088611477224235621321607$$ $$a = \log_{10}m = n^2 \log_{10}n=41.40980396069858470489859667103917348$$ $$10^f=2.569235775210588780886114772242356213216070000$$