How can you prove that $1.05^{50} < 100$ without a calculator? Is there a way to prove that $1.05^{50} &lt; 100$ without a calculator?

I have tried this...

$$1.05^{50}&lt;10^2$$
$$(\frac{105}{100})^{50}&lt;10^2$$
$$(\frac{21}{20})^{50}&lt;10^2$$
$$\frac{21^{50}}{20^{50}}&lt;10^2$$
$$\frac{21^{50}}{2^{50}*10^{50}}&lt;10^2$$
$$\frac{21^{50}}{2^{50}}&lt;10^{52}$$
$$10.5^{50}&lt;10^{52}$$

...but I don't know where to go. Can someone assist me (alternate methods are fine)?

EDIT:
 Can anyone help me prove that $2^{1000}&lt;10^{302}$ without a calculator?
$$
1.05^{50} = \left(1+\frac{5}{100}\right)^{50}
= \sqrt{\left(1+\frac{5}{100}\right)^{100}}
&lt; \sqrt{e^5} = \sqrt{e}^5 &lt; 2^5 = 32
$$

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



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

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

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

We want $\left(\frac{2^{10}}{10^3}\right)^{100}&gt;10$. From $(1-t)^a\leq e^{-at}$ you deduce $\left(\frac1{1-t}\right)^a\geq e^{at}$. So
$$
\begin{split}
\left(\frac{2^{10}}{10^3}\right)^{100}
&amp;= \left(\frac{1024}{1000}\right)^{100}
= \left(\frac{1}{1-\frac{24}{1024}}\right)^{100}
\geq e^{100\cdot\frac{24}{1024}} = e^{75/32} \\
&amp;= e^{2+11/32} &gt; e^{2+11/33}
= \exp(2)\exp(1/3) &gt; \exp_5(2)\exp_2(1/3) \\
&amp;= \left(1+2+2+\frac43+\frac23+\frac4{15}\right)\left(1+\frac13+\frac1{18}\right) \\
&amp;= \frac{109}{15} \cdot \frac{25}{18} = \frac{545}{54} &gt; 10.
\end{split}
$$
By writing 1.05 as (1 + 0.05) and expanding by using binomial Theorem, you can clearly see that in all terms there is positive power in 0.05 ( except the first one) which is making it smaller. So it can never be greater than 100 or even leave 100 it can never be greater than 50.