Dirac is a distribution and not a function. That means it can only be defined by an integral :
$$<\delta, \phi> = \int{\phi\cdot d\delta} = \phi(0)$$
Now if we take the Fourier transform of a Dirac, in a distribution sense :
$$ s(x) = \delta $$
$$
\begin{split}
S(f) &= \int^{+\infty}_{-\infty}{s(x)\times e^{-2\pi ifx}dx}\\
&=\int^{+\infty}_{-\infty}{e^{-2\pi ifx}d\delta}\\
&=e^{-2\pi if\times0} = 1
\end{split}
$$
Which means that reciprocally, $TF(1) = \delta$.
Considering that Fourier Transform is linear, $TF(C) = C\times \delta$
The other think to take in account is that a DC signal in the pure mathematical sense doesn't physically exist : it would mean that the signal has existed for an inifinite period of time and will continue to do so until the end of infinity.