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 $7^{2\cdot 15}\cdot 7 > 8^{2\cdot 14}\cdot 8$. I also tried to write the right hand side as $\frac{8^{31}}{8^2}$.

First, $7^5=16807$ and $2^{14}=16384$. And we have $\dfrac{7^{31}}{8^{29}}=\left(\dfrac{7^5}{2^{14}}\right)^6\times\dfrac78$.
Now, for $x\ge0$ and integer $n\ge0$, we have $(1+x)^n\ge1+nx$, hence
$$\left(\frac{7^5}{2^{14}}\right)^6\ge1+6\left(\frac{7^5}{2^{14}}-1\right)=1+\frac{6\times423}{2^{14}}=1+\frac{3\times423}{2^{13}}=\frac{9461}{8192}$$
Finally,
$$\frac{9461}{8192}\times\frac78=\frac{66227}{65536}>1$$
Therefore,
$$\frac{7^{31}}{8^{29}}>1$$
