Which one is bigger: $9^{17}$ and $7^{19}$

$$7^{19}=7\cdot49^9\lt7\cdot50^9=7\cdot125^3\cdot10^9\lt72\cdot128^3\cdot10^8=9\cdot2^{24}\cdot10^8=9\cdot80^8\lt9\cdot81^8=9^{17}$$

An alternative approach, assuming this is doable enough by hand: $$\color{blue}{\left(\frac{7}{9}\right)^3} = \frac{343}{729} \color{blue}{< \frac{1}{2}}$$ Then: $$\color{red}{\frac{7^{19}}{9^{17}}} = \left(\frac{7}{9}\right)^{18} \; \frac{9}{7} \; 7^2 = \left( \color{blue}{\left(\frac{7}{9}\right)^3} \right)^6 \; 63 \color{blue}{<} \left(\color{blue}{ \frac{1}{2}} \right)^6 63 = \frac{63}{64} \color{red}{< 1}$$ So $\color{red}{7^{19}<9^{17}}$ after fairly easy and straightforward calculations.