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

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$$ 3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}. $$

$$17 ^{ 13} = ((17^3)^2)^2 \cdot 17= (4913 \cdot 4913)^2\cdot 17< (242\cdot10^5)^2\cdot 17\\< 588\cdot10^{12}\cdot 17= 9996\cdot10^{12}<10^{16} $$

Hence, $$17 ^{ 69} = \left(17^{13}\right)^{\frac{69}{13}}<10^{16\cdot(5+\frac{4}{13})}= 10^{80+\frac{64}{13}} < 10^{85}.$$

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Here are some minor tricks to make the computation in the first inequalities even easier.

$$4913 \cdot4913=(4910+3)(4920-7) < 4910\cdot4920$$ $$491 \cdot492=(500-9)(500-8)= 241572$$ $$242\cdot242=(240+2)(245-3)<240\cdot245=12\cdot490= 58800$$

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Similar formulas:

$$\begin{array}{cl} \left.17^{4}\right/10^{5} &=0.83521\\ \left.17^{13}\right/10^{16} &=0.990458\!\cdots\\ \left.17^{69}\right/10^{85} &=0.796115\!\cdots\\ \left.17^{243}\right/10^{299} &=0.997902\!\cdots\\ \left.17^{1202}\right/10^{1479} &=0.999087\!\cdots\\ \left.17^{5524}\right/10^{6797} &=0.999636\!\cdots\\ \left.17^{7685}\right/10^{9456} &=0.999910\!\cdots\\ \vdots\\ \left.17^{302464054}\right/10^{372166569} &=0.99999999988\cdots\\ \end{array}$$

The above data is generated with, among other tools, the continued fraction of $$ \log_{17}10= 0.81271150929195899925562198972659\cdots,$$ which is, $$ [0; 1, 4, 2, 1, 17, 1, 3, 1, 1, 3, 3, 26, 1, 1, 2, 3, 2, 11, 64, 2, 3, 1, 13, 1, 8, 1, 4, \cdots].$$

Claim 1: $2.3<\ln 10.$

Claim 2: $\ln 1.7<8/15$

Both these claims can be proven easily via Taylor series, etc.

Now, using the above inequalities, we have $1.7^{69}<e^{69\cdot \frac{8}{15}}<10^{16},$ or, multiplying $10^{69}$ on both sides, $17^{69}<10^{85}.$