How should I interpret the sampling theorem? I recently learned the following:

Theorem: If $f(t)$ is a band-limited signal with cut-off frequency $\Omega/2\pi$, then
  $$
f(t) = \sum_{n\in\mathbb{Z}} f\left(\frac{n\pi}{\Omega}\right) \operatorname{sinc}(\Omega t - n\pi)
$$

Of all the things I've seen in Fourier analysis, this is by far the strangest. For some reason, I'm having trouble deriving any substantial meaning from this formula, other than the fact that $f$ can be "recovered" from its values at certain intervals.
I really don't understand this intuitively like I do other things, like the Fourier series or Fourier transform. Why can we do this if $f$ is band-limited, and why would "band-limitedness" be the condition that allows this to occur? How should I interpret the $\text{sinc}$'s? This is all just very confusing and meaningless to me.
 A: Suppose we sample $f$ at the points $x_n = n T$. How can we recover the full function $f$, given only these sample values? 
The key idea (a brilliant idea) is to recognize that sampling $f$ at these particular values is equivalent to multiplying $f$ by a Dirac comb function $g_T$ which has an infinitely sharp spike of unit area at each point $x_n$. Here's a visualization of a Dirac comb from Wikipedia: 
$\qquad \qquad$
Let
$$
h = g_T \cdot f.
$$
By "equivalent", I mean that knowing the values of $f$ at the points $x_n$ is equivalent to knowing the function $h$.
Of course, the most important fact about the Fourier transform is that multiplication in the time domain corresponds to convolution in the frequency domain. Moreover, it can be shown that the Fourier transform of a comb function is a comb function. Thus, 
$$
\hat h = \hat g_T * \hat f = \frac{1}{T} g_{\frac{1}{T}} * \hat f.
$$
What happens when you convolve with a comb function, as we are doing here?  If you visualize the graph of $g_{\frac{1}{T}} * \hat f$, it consists of a bunch of shifted copies of the graph of $\hat f$. If we assume that $f$ is sufficiently band-limited, then crucially these copies do not overlap, which is very nice because we are now able to recover $\hat f$ exactly by multiplying $\hat h$ by a "rectangular function" $\hat r$:
$$
\hat f = \hat r \cdot \left( g_{\frac{1}{T}} * f \right) = \hat r \cdot (T \hat h).
$$
Here's a picture from Wikipedia that illustrates this idea nicely:

Now we are ready to recover $f$ by taking the inverse Fourier transform of both sides:
$$
\tag{$\spadesuit$} f = r * (T h).
$$
We can now see where the sinc function comes from, because the inverse Fourier transform of the rectangular function $\hat r$ is the sinc function. Equation ($\spadesuit$) expresses $f$ as a sum of shifted copies of the sinc function, with each copy weighted by the corresponding sample value of $f$. Written out explicitly, we have discovered the formula for $f(t)$ given above in the question statement.
