For this particular example, it's pretty easy, since
$$
(9/10)^8 \geq (8/10)^4 \geq (6/10)^2 \geq (3/10)
$$
and so you're done. Notice, my bounds are quite crude in the last two cases.
Repetitive squaring is the quickest way to get up to a large power. In fact, that's how a typical computer implementation will do it for integer powers.
Consider $f(x,y) = x^y - y^x$ where $x, y \approx 3$. Increasing $x$ should bring $f$ down and increasing $y$ should bring $f$ up. In general, without any real knowledge of $f$:
$$ 1 > f(x+ \epsilon_1, y + \epsilon_2) \approx
f(x, y ) + \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
$$
where the first inequality is something we are trying to prove for $x = e \approx 2.718$ and $y = \pi \approx 3.141$.
Let's try $x = \frac{11}{4}$ and $y = \frac{13}{4}$. Then we have taken off too much slack:
$$ \left( \frac{11}{4} \right)^{\frac{13}{4}} - \left( \frac{13}{4} \right)^{\frac{11}{4}} \approx 1.214$$
Using the continued fractions of $e$ and $\pi $ does give less than one and check with a calculator:
$$ \left( \frac{8}{3} \right)^{\frac{22}{7}} - \left( \frac{22}{7} \right)^{\frac{8}{3}} \approx 0.621$$
Even your proof requires a calculator in the last step... but we did not use the exact values of $\pi, e$.
How good is this approximation? Somehow we should compute how quickly $f$ changes with $x$ and $y$:
$$ \frac{\partial f}{\partial x} = y \, x^{y-1} - (\ln y ) y^x \approx 3 \times 3^{3-1} - (\ln 3) 3^3 \approx -27(\ln \frac{e}{3}) \approx -2.7 $$
The $y$ partial derivative is similar, except it is positive instead of negative.
It can be proven, the continued fraction error of $e$ and $\pi$ are inversely proportional to the denominators:
$$ \left| e - \frac{8}{3} \right| < \frac{1}{3^2} \text{ and } \left| \pi - \frac{22}{7} \right| < \frac{1}{7^2}$$
These would be our estimates of $\epsilon_1$ and $\epsilon_2$. Then using a tiny bit of multivariable calculus:
$$ \epsilon_1 \frac{\partial f}{\partial x} + \epsilon_2 \frac{\partial f}{\partial y}
\approx \frac{1}{3^2} \times 2.7 + \frac{1}{7^2} \times (-2.7) < 0.355 $$
This should be enough to establish $|f(\pi,e)| < 1$.
This used a calculator in many places but not the exact value of the constants $\pi$ and $e$.
Best Answer
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$$ 3^{23} = (3^9)^2 × 3^5 < (2 × 10^4)^2 × (25 × 10) = 10^{11}. $$