TITLE: A simpler non-calculator proof for $17^{69}<10^{85}$
QUESTION [5 upvotes]: I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.
My proof is the following:
\begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}
Could anyone find a simpler non-calculator proof without using big numbers?

REPLY [2 votes]: I will also say some words on this. The general procedure to easily show such inequalities without computer is to... use the computer to get "close powers" of the bases, here $17$ and $10$, then use the coarsest that still does the job, and of course not mention that this was done so! In our case, i am forgetting in this second to not mention that the "first closest powers" of $17$ and $10$ come from the convergents of the continued fraction of $a=\displaystyle\log_{10} 17=\frac {\log 17}{\log 10}$, so let us show them...
sage: c = continued_fraction( log(17)/log(10) )
sage: cvgts = [ c.convergent(k) for k in [1..7] ]
sage: cvgts
[5/4, 11/9, 16/13, 283/230, 299/243, 1180/959, 1479/1202]

So we expect that 

$17^4=83521$ is "close" to $10^5$, yes, this is the case and $17^4\color{blue}{<}10^5$,
$17^9=118587876497$ is "close" to $10^{11}$, yes, and $17^{9}\color{red}{>}10^{11}$,
$17^{13}=9904578032905937$ is "close" to $10^{16}$, yes, and $17^{13}\color{blue}{<}10^{16}$,
$17^{230}=\dots$ is "close" to $10^{283}$, yes, and $17^{230}\color{red}{>}10^{283}$, and so on.

Now we "completely forget" about the above, and write some inequalities. I will use the knowledge of the "steps" $17^4$, and $17^{13}$ below (of course, without mentioning this)... It will be a "hard job" (more than four lines) to establish $17^{13}\le 10^{16}$, but then we can relax and easily show the needed inequality. The most complicated operation will be to compute $836^2$ below. So let us start now!

$$
\begin{aligned}
17^4 &= 289^2 = (290-1)^2 = 84100-580+1=83521\\
 &< 83600\ ,
\\[3mm]
%
17^{13} &=17\cdot (17^4)^3\\
 &< 17\cdot 836^3\cdot 10^6 \\
 &= 17\cdot 836^2\cdot 836\cdot 10^6 \\
 &=   17\cdot 698896\cdot 836\cdot 10^6 \\
 &< 17\cdot 700000\cdot 840\cdot 10^6 \\
 &= 17\cdot 7\cdot 84\cdot 10^{12} \\
 &= 9996\cdot 10 ^{12}\\
 &< 10^{16}\ ,
\\[3mm]
%
17^{69} &= (17^{13})^5\cdot 17^4\\
 &<(10^{16})^5\cdot 83600\\
 &< 10^{80}\cdot 10^5\\
 &= 10^{85}\ .
\end{aligned}
$$