I think it's worth seeing a proof that is based, more or less, on first principles.
Let $\alpha$ be an irrational number, and let $f(x)=x^\alpha$, for $x\gt0$. By the definition of derivative,
$$f'(x)=\lim_{t\to x}{f(t)-f(x)\over t-x}=\lim_{t\to x}{t^\alpha-x^\alpha\over t-x}=\lim_{u\to1}{(xu)^\alpha-x^\alpha\over xu- x}=x^{\alpha-1}\lim_{u\to1}{u^\alpha-1\over u-1}$$
so it suffices to show
$$\lim_{u\to1}{u^\alpha-1\over u-1}=\alpha$$
It will turn out to be helpful to reduce things to computing a one-sided limit. To do so, note that
$$\lim_{u\to1^-}{u^\alpha-1\over u-1}=\lim_{v\to1^+}{v^{-\alpha}-1\over v^{-1}-1}=\lim_{v\to1^+}\left(v^{1-\alpha}{v^\alpha-1\over v-1}\right)=\lim_{v\to1^+}{v^\alpha-1\over v-1}$$
since $\lim_{v\to1}v^p=1$ for any power $p$. In sum, changing notation slightly, we need to show
$$\lim_{h\to0^+}{(1+h)^\alpha-1\over h}=\alpha$$
That is, we need to show that for any $\epsilon\gt0$ there exists $\delta\gt0$ such that
$$0\lt h\lt\delta\implies\left|{(1+h)^\alpha-1\over h}-\alpha\right|\lt\epsilon$$
Now let's assume we know that the derivative of $x^r$ is $rx^{r-1}$ when $r$ is rational. Furthermore, let's assume we know the generalized Bernoulli inequality, $(1+h)^p\le1+ph$ for $0\lt p\lt1$. With this in hand, let $r$ be a rational number slightly larger than $\alpha$ (how slight, we'll see momentarily). Then
$$\begin{align}
{(1+h)^\alpha-1\over h}-\alpha
&=\left({(1+h)^\alpha-1\over h}-{(1+h)^r-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\\
&=-(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\\
\end{align}$$
Given $\epsilon\gt0$, we can clearly choose $r$ so that $|r-\alpha|\lt{\epsilon\over3}$. Once we fix that choice of $r$, the known derivative formula for $x^r$ means we can choose a $\delta_r$ such that
$$0\lt h\lt\delta_r\implies\left|{(1+h)^r-1\over h}-r\right|\lt{\epsilon\over3}$$
Finally, we see that, if $0\lt h\lt1$ and $0\lt\alpha\lt r$, then
$$0\lt(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)\le(1+h)^\alpha\left({(1+(r-\alpha)h)-1\over h}\right)=2^\alpha(\alpha-r)$$
So let's choose $r\lt\alpha$ so that $2^\alpha(\alpha-r)\lt{\epsilon\over3}$ and then let $\delta=\min\{\delta_r,1\}$. Then we have
$$\begin{align}
0\lt h\lt\delta\implies\left|{(1+h)^\alpha-1\over h}-\alpha\right|
&=\left|-(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)+\left({(1+h)^r-1\over h}-r\right)+(r-\alpha)\right|\\
&\le\left|(1+h)^\alpha\left({(1+h)^{r-\alpha}-1\over h}\right)\right|+\left|{(1+h)^r-1\over h}-r\right|+\left|r-\alpha\right|\\
&\lt{\epsilon\over3}+{\epsilon\over3}+{\epsilon\over3}=\epsilon
\end{align}$$
as desired.
Remark: Bernoulli's inequality, $(1+h)^p\le1+ph$ if $0\lt p\lt 1$ and $h\gt0$ can be proved by differentiating the function $f(h)=(1+h)^p-ph$ and showing that $f$ is strictly decreasing from its value $f(0)=1$. This may seem circular, since we're applying the inequality to the irrational power $r-\alpha$. But we really only need the derivative-based proof for rational exponents $p$. Then, if $0\lt\beta\lt1$ is an irrational number, we have
$$(1+h)^\beta\lt(1+h)^p\le1+ph$$
for any rational $p\in(\beta,1)$, after which it suffices to let $p\to\beta$.