Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ 
Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.

I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 \cdot 1976^7>0$. How should I do that?

$$E=\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}-\frac{1}{10^{1444}}\gt 0\large?$$
$$E=1-\frac{1}{4\cdot 1976^4}-\frac{4\cdot 1976^3}{10^{1444}}\gt 0\large?$$
The answer is YES because $$\frac{1}{4\cdot 1976^3}\lt \frac 14$$ and because the numerator $4\cdot 1976^3$ has $11$ digits.
