A slightly clearer but somewhat less rigorous connection (only C'bility \Longrightarrow rank condition!) can be made as follows: We can solve the diff. eq. system that you have provided and obtain
<br />
x(t) = \int_0^{\infty}e^{A(t-\tau)}Bu(\tau)d\tau + e^{At}x(0)<br />
Let's assume zero initial conditions for simplicity. Now, since the controllability means that I can reach any x(t), the integral converges to x(t) with some u(t). Let's use the Taylor series of exponential
<br />
x(t) = \int_0^{\infty}\left(I+A(t-\tau) + \frac{A^2}{2!}(t-\tau)^2+\cdots \right)Bu(\tau)d\tau<br />
You can take the constant terms out and obtain a matrix-vector multiplication (though infinite dimensional)
<br />
x(t) = \begin{bmatrix}B &AB &A^2B &\cdots\end{bmatrix}\begin{pmatrix}\int_0^{\infty}u(\tau)d\tau \\\int_0^{\infty}(t-\tau)u(\tau)d\tau \\ \int_0^{\infty}\frac{1}{2!}(t-\tau)^2u(\tau)d\tau\\ \vdots\end{pmatrix} = \mathcal{C}_\infty \mathcal{U}<br />
I would denote the matrix part as \mathcal{C}_\infty . Now, since we assume controllability, we should be able to obtain any x(t), hence \mathcal{C}_\infty must be full row rank. But from Cayley-Hamilton theorem we know that the powers of A with degree higher then n-1, can be rewritten by the powers of A up to the degree n-1. (This is a bad sentence but looking it up is easy so I skip that part.) This means that no extra information about the rank of this matrix can be included after the A^{n-1}B since the remaining terms are linear combinations of the first n terms. Thus,
rank(\mathcal{C}_\infty) = rank(\mathcal{C}) = rank(\begin{bmatrix}B &AB &A^2B &\cdots &A^{n-1}B\end{bmatrix}
In case of SISO systems, \mathcal{C} happens to be square so the rank condition equals to the determinant being nonzero.
