A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}&lt;10^{85}$ by using the following inequalities:
$x&lt;\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x&lt;{\mathrm e}^{x-1}$ 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}}&amp;=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}&lt;\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}\\&amp;&lt;\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)\\&amp;=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)&lt;1.\end{align*}

Could anyone find a simpler non-calculator proof without using big numbers?
Since $17^3 = 4913 &lt; 492 × 10$, then$$
17^6 &lt; 492^2 × 10^2 = 242064 × 10^2 &lt; 243000 × 10^2 = 3^5 × 10^5.
$$
Now it suffices to prove that $(3^5 × 10^5)^{23} &lt; (10^{85})^2$, or $3^{23} &lt; 10^{11}$. Note that $3^9 = 27^3 = 19683 &lt; 2 × 10^4$ and $3^5 = 243 &lt; 25 × 10$, thus$$
3^{23} = (3^9)^2 × 3^5 &lt; (2 × 10^4)^2 × (25 × 10) = 10^{11}. 
$$
Proof that $17^{69} &lt; 10^{85}$ without using calculator, there are many ways this can be done you'll need to understand number theory, so I'll highlight a few examples
$$17^{69} &lt; 10^{85}$$
$$17^{69} &lt; 10^{17×5}$$
$$17^{1/17} &lt; 10^{5/69}$$
Remember $$\lim_{n→∞} \sqrt[n]{n} = 1$$
Meaning the value of 17^{1/17} is between $1$ to $1.4$
$$17^{1/17} &lt; 10^{5/69}$$
Using long division $\frac{5}{69} = 0.07246$
$$17^{1/17} &lt; 10^{x}$$
But $10^x$ can also be between $1$ to $1.4$ if $x$ is from $0$ to $≈0.1$
So the the value $\frac{5}{69}$ is too small, therefore
$$17^{69} \mathbb{&lt;} 10^{85}$$

Another way
$$17^{69} &lt; 10^{85}$$
$$17^{69} &lt; 10^{17×5}$$
$$17^{69/17} &lt; 10^{5}$$
Using long division $\frac{69}{17} = 4.0588$
$$17^{4.0588} &lt; 10^{5}$$
If we assume that $17^{4.0588} = 20^x$, since the base became bigger, the power would be smaller for them to be equal

$17 → 20$$, $4.0588 → ≈3$

$$17^{4.0588} ≈ 20^3$$
$$≈20^3 &lt; 10^5$$
But we know clearly that $8000 &lt; 10000$
Therefore 
$$17^{69} \mathbb{&lt;} 10^{85}$$