Prove that $2^{2^{\sqrt3}}>10$ With a computer or calculator, it is easy to show that
$$
2^{2^\sqrt{3}} = 10.000478 \ldots &gt; 10.
$$

How can we prove that $2^{2^{\sqrt3}}&gt;10$ without a calculator?
I suspect any adequate answer to this question is going to be very computation-heavy.  Here's one.

First we claim that
$$
\sqrt{3} &gt; \frac{3691}{2131} \tag{1}
$$
This fraction is not pulled out of a hat; rather, it is found by taking the continued fraction $\sqrt{3} = [1; 1, 2, 1, 2, 1, 2, \ldots]$ and carrying it out a while.  Anyway, the above is proved directly by
$$
3 \cdot 2131^2 = 3 \cdot 4541161 = 13623483 &gt; 13623481 = 3691^2.
$$

We next claim that
$$
2^{{3691}/{2131}} &gt; \frac{1465}{441} \tag{2}
$$

To prove this, one needs to show $2^{3691} 441^{2131} &gt; 1465^{2131}$.
With repeated squaring, each term requires squaring about $12$ times, so this is doable$^{a}$ with a pencil and paper.

The last step is
$$
2^{1465/441} &gt; 10 \tag{3}
$$
Again one uses repeated squaring, and if one is working in base 10 one only needs to count the number of digits in $2^{1465}$.  This should not take nearly as long as the previous computation.$^{b}$

From (1), (2), and (3),
$$
2^{2^{\sqrt{3}}} &gt; 2^{2^{3691/2131}} &gt; 2^{1465 / 441} &gt; 10. \quad \square
$$



$^{a}$ Honestly, this is quite a stretch. I can't guarantee it won't take like a year of work.

$^{b}$ As a rough estimate, perhaps a day or so.
First take logs
$$2^{\sqrt3}&gt;\log_2 10$$
Then again
$$\sqrt{3}&gt;\log_2 \log_2 10$$
Now, $$\log_2(10)=\frac{\log 10}{\log 2}$$
With $\log 10\approx 2.302$ and $\log 2\approx 0.6931$, We find $$\log_2(10)\approx3.32131$$ 
And the $\log_2$ of that is $(\approx1.731)$ Now, $\sqrt 3\approx 1.732,$ and therefore we conclude that $$2^{2^{\sqrt3}}&gt;10$$
Note that these values can be found with Newton's method and taylor series, though tedious to find and calculate.