Problem source

Let $x=10^{100}$,  $y=10^{x}$,  $z=10^{y}$, 
and let
$$ 
a_1=x!, \quad a_2=x^y,\quad a_3=y^x,\quad a_4=z^x,\quad 
   a_5=\e^{xyz},\quad a_6=z^{1/y},\quad a_7 = y^{z/x}.
$$
\begin{questionparts}
\item Use Stirling's approximation 
$n! \approx \sqrt{2 \pi}\, {n^{n+{1\over2}}\e^{-n}}$, which is valid for 
large $n$, to show that 
$\log_{10}\left(\log_{10} a_1 \right) \approx 102$.
\item Arrange the seven numbers $a_1$, $\ldots$ , $a_7$ in ascending 
order of magnitude, justifying  your result.
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
\log_{10}(\log_{10} a_1) &= \log_{10} (\log_{10} (x!) \\
&\approx \log_{10} (\log_{10} \sqrt{2 \pi} x^{x+\frac12} e^{-x}) \\
&= \log_{10} \l \log_{10} \sqrt{2 \pi} + (x+\frac12) \log_{10} x-x \r \\
&= \log_{10} \l \log_{10} \sqrt{2 \pi} + (100x+50)-x \r \\
&= \log_{10} \l 99x + \epsilon \r \\
&\approx \log_{10} 99 + \log_{10} x \\
&\approx 2 + 100 = 102
\end{align*}

\item 
\begin{align*}
\log_{10}(\log_{10} a_2) &= \log_{10}(\log_{10} x^y) \\ &= \log_{10} y + \log_{10} \log_{10} x \\
&= x + 2
\end{align*}

\begin{align*}
\log_{10}(\log_{10} a_3) &= \log_{10}(\log_{10} y^x) \\ &= \log_{10} x + \log_{10} \log_{10} y \\
&= 100 + \log_{10} x \\
&= 200
\end{align*}

\begin{align*}
\log_{10}(\log_{10} a_4) &= \log_{10}(\log_{10} z^x) \\ &= \log_{10} x + \log_{10} \log_{10} z \\
&= 100 + \log_{10} y \\
&= 100+x
\end{align*}

\begin{align*}
\log_{10}(\log_{10} a_5) &= \log_{10}(\log_{10} e^{xyz}) \\ 
&= \log_{10} x + \log_{10}y+\log_{10} z+ \log_{10} \log_{10} e \\
&\approx 100 + x + y 
\end{align*}

\begin{align*}
\log_{10}(\log_{10} a_6) &= \log_{10}(\log_{10} z^{1/y}) \\
&= \log_{10}(\log_{10} 10) \\
&= 0
\end{align*}

\begin{align*}
\log_{10}(\log_{10} a_7) &= \log_{10}(\log_{10} y^{z/x}) \\
&= \log_{10}z-\log_{10} x + \log_{10} \log_{10} y \\
&= y 
\end{align*}

Since $0 < 102 < 200 < x+2 < x+100 < y < y+x+100$ we must have

$a_6 < a_1 < a_3 < a_2 < a_4 < a_7 < a_5$
\end{questionparts}