How to prove that $7^{31} > 8^{29}$ How can I prove that $7^{31}$ is bigger than $8^{29}$?

I tried to write exponents as multiplication, $2\cdot 15 + 1$, and $2\cdot 14+1$, then to write this inequality as $7^{2\cdot 15}\cdot 7 &gt; 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.
The following (not particularly elegant) proof uses reasonably basic multiplication and division.

We need to show that $7^{31} &gt; 8^{29}$, i.e. that $\dfrac{7^{31}}{8^{29}}&gt;1$.

We have: $\dfrac{7^{31}}{8^{29}}=\dfrac{7^{2}\cdot7^{29}}{8^{29}}=\dfrac{7^{3}}{8}\Big(\dfrac{7}{8}\Big)^{28}=\dfrac{7^{3}}{8}\Big(\dfrac{7^4}{8^4}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{2401}{4096}\Big)^{7} &gt; \dfrac{7^{3}}{8}\Big(\dfrac{2400}{4100}\Big)^{7}=\dfrac{7^{3}}{8}\Big(\dfrac{24}{41}\Big)^{7}=\dfrac{7^{3}}{8}\dfrac{24}{41}\Big(\dfrac{24}{41}\Big)^{6}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{24^2}{41^2}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1681}\Big)^{3}&gt;\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{576}{1683}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{9 \cdot 64} {9\cdot 187}\Big)^{3}=\dfrac{3 \cdot 7^{3}}{41}\Big(\dfrac{64} {187}\Big)^{3}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{11^3 \cdot 17^3 \cdot 41}=\dfrac{2^{18} \cdot 3 \cdot 7^3}{1331 \cdot 17^3 \cdot 41}&gt;\dfrac{2^{18} \cdot 3 \cdot 7^3}{1332 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{111 \cdot 17^3 \cdot 41}=\dfrac{2^{16} \cdot 7^3}{4551 \cdot 17^3}&gt;\dfrac{2^{16} \cdot 7^3}{4557 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{93 \cdot 17^3}=\dfrac{2^{16} \cdot 7}{1581 \cdot 17^2}&gt;\dfrac{2^{16} \cdot 7}{1582 \cdot 17^2}=\dfrac{2^{15}}{113 \cdot 17^2}=\dfrac{2^{15}}{1921 \cdot 17}&gt;\dfrac{2^{15}}{1924 \cdot 17}=\dfrac{2^{13}}{481 \cdot 17}=\dfrac{8192}{ 8177}&gt;1. \quad\square$
I tried just approximate the numbers and I think I found a very ugly way to prove the inequality. Are there any mistakes?

We have $7^4&gt;24\cdot10^2$ so $7^{31}=7^3\cdot(7^4)^7&gt;7^3\cdot(24\cdot10^2)^7=2^{35} \cdot 3^7 \cdot 5^{14} 
\cdot 7^3$. Also
$8^{29}=2^{87}$. So one has to prove the inequalities $2^{52}&lt;3^7 \cdot 5^{14} \cdot 
7^3=750141\cdot5^{14}&lt;750142\cdot5^{14}$ or $2^{51}&lt;375071\cdot5^{14}&lt;375072\cdot5^{14}$.

This mean we have to 
prove that $2^{51}&lt;2^5 \cdot 3 \cdot 3907\cdot5^{14}$ or 
$2^{46}&lt;3\cdot3907\cdot5^{14}&lt;3\cdot3908\cdot5^{14}=3\cdot2^2\cdot977\cdot5^{14}$ or equivalently
$2^{44}&lt;3\cdot977\cdot5^{14}$. But $3\cdot977\cdot5^{14}&lt;3\cdot976\cdot5^{14}=3\cdot2^4\cdot61\cdot5^{14}$ so 
one has to prove that $2^{40}&lt;3\cdot61\cdot5^{14}$. 

But $3\cdot61\cdot5^{14}&lt;3\cdot62\cdot5^{14}$ so one has to 
prove that $2^{40}&lt;3\cdot62\cdot5^{14}$ or $2^{39}&lt;3\cdot31\cdot5^{14}$. But 
$3\cdot31\cdot5^{14}&lt;3\cdot2^5\cdot5^{14}$ so one has to prove that $2^{34}&lt;3\cdot5^{14}$.

Now, lets approximate some square roots:

The inequality above is equivalent to $2^{17}&lt;\sqrt{3}\cdot5^7$. But $\sqrt{3}&gt;1.7$ so one has to prove that $2^{17}&lt;1.7\cdot5^7$. This is the same as $\sqrt{2}\cdot2^8&lt;\sqrt{1.7\cdot5}\cdot5^3$ or $256\cdot\sqrt{2}&lt;\sqrt{8.5}\cdot125$. This is 
the same as $(256/125)^2&lt;8.5/2$ or $256^2\cdot2&lt;8.5\cdot125^2$ or $131072&lt;132812.5$.