How to sample $f:=f_{1}\cdot f_{2}\cdot...\cdot f_{K}$?
The definition of bandlimited is that $\hat f_k$ has compact support of size $\sigma_k$. To make our argument easier, we'll also note that we can decompose every $\hat f_k$ into $\tilde f_k * \delta_k$ (a convolution with a dirac impulse, just a shift) so that $\tilde f_k$ is "0-aligned", ie. $\sup f_k = \sigma_k$ and $\inf f_k=0$.

For $f_1\cdot f_2$ follows directly that the support of $\tilde f_1*\tilde f_2$ has the property

$$\text{supp}(f_1*f_2) \subset [-\sigma_2;\sigma_1]$$

from the way we calculate the convolution:

$$(f_1*f_2)(x)= \int\limits_{-\infty}^\infty f_1(y)f_2(x-y)\,dy$$

Now, that support has "length" (to be more exact, an upper boundary for measure) $\sigma_1+\sigma_2$.

It's very intuitive to now apply induction to show that this applies to all $k$. 

So, in the end, $\text{supp}\left({\Large*}\left(\tilde f_k\right)_{k=1}^{K}\right)\subset [0, \sum\limits_{k=1}^K \sigma_k]$.

Since the shifting around in frequency domain doesn't theoretically (measure of $\delta_k0\,\forall k$) nor practically matter (compare: undersampling), we can just sample sufficiently by observing the sum bandwidth.

Note that this has very practical effects, for example, when sampling signals that are reflected by moving targets.