The defining formula
$$\mathcal{F}f(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) \, e^{-ikx} \, dx$$
only works when $f \in L^1(\mathbb{R}).$ The Fourier transform can however be extended to several other cases. For example, for a tempered distribution $u$ it is defined to be the distribution $\mathcal{F}u$ satisfying
$$\int_{-\infty}^{\infty} \mathcal{F}u(x) \, \varphi(x) \, dx = \int_{-\infty}^{\infty} u(x) \, \mathcal{F}\varphi(x) \, dx,$$
for all test functions $\varphi$ in Schwartz space $\mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R}).$
One example of a tempered distribution is $\delta.$ Its Fourier transform is thus given by
$$
\int_{-\infty}^{\infty} \mathcal{F}\delta(x) \, \varphi(x) \, dx
= \int_{-\infty}^{\infty} \delta(x) \, \mathcal{F}\varphi(x) \, dx
= \mathcal{F}(0)
= \frac{1}{\sqrt{2\pi}} \left. \int_{-\infty}^{\infty} \varphi(x) \, e^{-ikx} \, dx \right|_{k=0}
= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} \varphi(x) \, dx
,
$$
i.e.
$\mathcal{F}\delta = \frac{1}{\sqrt{2\pi}}$ (constant function).
For $\varphi \in \mathcal{S}(\mathbb{R})$ we have $\mathcal{F}^2\varphi(x) = \varphi(-x),$ and it's easy to show from this that also for tempered distributions we have $\mathcal{F}^2 u(x) = u(-x).$ Therefore, for the constant function $1$ we have
$$
\mathcal{F}1(x)
= \mathcal{F}\{\sqrt{2\pi}\,\mathcal{F}\delta\}(x)
= \sqrt{2\pi}\,\mathcal{F}^2\delta(x)
= \sqrt{2\pi}\,\delta(-x)
= \sqrt{2\pi}\,\delta(x).
$$
Best Answer
Using formal induction
We want to show that $\mathcal{F}\{f^{(n)}(x)\} = (ik)^n \, \mathcal{F}\{f(x)\}.$
Basecase: For $n=0$ the identity is trivial.
Induction step: Assume that the identity is valid for some integer $n=p \geq 0$. Then, $$\begin{align} \mathcal{F}\{f^{(p+1)}(x)\} &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f^{(p+1)}(x) \, e^{-ikx} \, dx \\ &= \frac{1}{\sqrt{2\pi}} \left( \left[f^{(p)}(x) \, e^{-ikx}\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} f^{(p)}(x) \, (-ik)e^{-ikx} \, dx \right) \\ &= ik \, \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f^{(p)}(x) \, e^{-ikx} \, dx \\ &= ik \, \mathcal{F}\{f^{(p)}(x)\} \\ &= ik \, \left((ik)^p \, \mathcal{F}\{f(x)\} \right) \\ &= (ik)^{p+1} \, \mathcal{F}\{f(x)\}, \\ \end{align}$$ i.e. it's also valid for $n=p+1.$
Conclusion: By the induction principle, the identity is valid for all integer $n \geq 0.$