A simpler non-calculator proof for $17^{69}<10^{85}$ 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?

Since $17^3 = 4913 < 492 × 10$, then$$
17^6 < 492^2 × 10^2 = 242064 × 10^2 < 243000 × 10^2 = 3^5 × 10^5.
$$
Now it suffices to prove that $(3^5 × 10^5)^{23} < (10^{85})^2$, or $3^{23} < 10^{11}$. Note that $3^9 = 27^3 = 19683 < 2 × 10^4$ and $3^5 = 243 < 25 × 10$, thus$$
3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}. 
$$
