How can you prove that $1.05^{50} < 100$ without a calculator?

$$ 1.05^{50} = \left(1+\frac{5}{100}\right)^{50} = \sqrt{\left(1+\frac{5}{100}\right)^{100}} < \sqrt{e^5} = \sqrt{e}^5 < 2^5 = 32 $$

Edit: for you second inequality, $2^{1000}<10^{302}$ is equivalent to $(2^{10})^{100}<10^2(10^3)^{100}$, which is equivalent to $$ \left(\frac{2^{10}}{10^3}\right)^{100} = \left(1+\frac{24}{1000}\right)^{100} < 100. $$ From $a\log(1+t)\leq at$ for $a>0$ we deduce $(1+t)^a\leq e^{at}$. Therefore $$ \left(1+\frac{24}{1000}\right)^{100} < e^{\frac{24}{1000}\cdot100} <e^{5/2} < 32 < 100. $$

---

Addendum. What is trickier, is proving that $2^{1000}>10^{301}$. Can you do that?

Here is how I go about it. Maybe someone else can find a simpler derivation.

Define $$ \exp_n(x) = \sum_{k=0}^n \frac{x^k}{k!} < \exp(x) . $$

We want $\left(\frac{2^{10}}{10^3}\right)^{100}>10$. From $(1-t)^a\leq e^{-at}$ you deduce $\left(\frac1{1-t}\right)^a\geq e^{at}$. So $$ \begin{split} \left(\frac{2^{10}}{10^3}\right)^{100} &= \left(\frac{1024}{1000}\right)^{100} = \left(\frac{1}{1-\frac{24}{1024}}\right)^{100} \geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\ &= e^{2+11/32} > e^{2+11/33} = \exp(2)\exp(1/3) > \exp_5(2)\exp_2(1/3) \\ &= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\ &= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} > 10. \end{split} $$

$\log(1.05^{50}) = 50\log(1.05) < 50·0.05 = 2.5 < 3$. Then $1.05^{50} < e^3 < 3^3 = 27$.

Here is an elementary way using GM-HM.

First note

• $1.05^{50} < 100 \Leftrightarrow \boxed{1.05 < \sqrt[50]{100}}$

$$\color{blue}{\sqrt[50]{100}} = \sqrt[50]{2^2\cdot 5^2 \cdot 1^{46}} \color{blue}{\stackrel{\mbox{GM-HM}}{>}}\frac{50}{\frac{2}{2}+\frac{2}{5}+46}=\frac{250}{237}=1+\frac{13}{237}> \color{blue}{1.05}$$