Hans Georg Schaathun
August 2016
$$f(x) = 2^x$$
$$ \begin{alignat*}{2} a^0 & = 1 && \\ a^n & = a\cdot a\cdot\ldots\cdot a \quad & n&=1,2,3,\ldots \\ a^{-n} & = \frac1{a^n} & n&=1,2,3,\ldots \\ a^{\frac mn} & = \sqrt[n]{a^m} & n&=1,2,3,\ldots, m&=\pm1,\pm2,\ldots \end{alignat*} $$
$$a^x = \mathop{\lim_{r\to x}}_{r\; \mathrm{rational}} a^r$$
$$2^\pi = $$
$$ \begin{align} r_1 &= 3 \\ r_2 &= 3{,}1 \\ r_3 &= 3{,}14 \\ r_4 &= 3{,}141 \\ r_5 &= 3{,}1415 \\ r_6 &= 3{,}14159 \\ & \vdots \end{align} $$
$$\pi = \lim_{n\to\infty} r_n$$
$$2^\pi = \lim_{n\to\infty} 2^{r_n}$$
$$ \begin{align} 2^3 &= 8 \\ 2^{3{,}1}&\approx 8{,}574187700290345 \\ 2^{3{,}14}&\approx 8{,}815240927012887 \\ 2^{3{,}141}&\approx 8{,}821353304551304 \\ 2^{3{,}1415}&\approx 8{,}824411082479122 \\ 2^{3{,}14159}&\approx 8{,}824961595059897 \\ & \vdots \end{align} $$
$$\lim_{n\to\infty} 2^{r_n} = 8{,}824\ldots$$