What is the maximum dynamic range of an N bit signal? In this presentation on slide 15 there is a corollary:


  Rule of thumb: the dark noise must be larger than 0.5
  
  Corollary: With a N bit digital signal you can deliver no more*) than
  N+1 bit dynamic range.
  
  *) You can if you use loss-less compression


They give an example:


  Example : A 102f camera with 11 bit dynamic range will deliver only 9
  bit in Mono8 mode. Use Mono16!


Why N+1? 
How can an 8 bit signal deliver 9 bit dynamic range?
From taking a look at the presentation, it seems that they might be using a slightly different definition of dynamic range than is typical. Usually, as in the Wikipedia article on the topic, dynamic range is defined as:

$$
\text{dynamic range} = \frac{\text{largest possible representable value}}{\text{smallest possible representable value}}
$$

For an $N$-bit (unsigned) signal, this is equal to:

$$
\text{dynamic range} = \frac{2^N-1}{1} = 2^N-1
$$

However, their discussion of dynamic range is interspersed with discussion of quantization noise. Therefore, I posit that they instead define dynamic range as:

$$
\text{dynamic range} = \frac{\text{largest possible representable value}}{\text{largest possible quantization error}}
$$

For a uniformly-quantized quantity such as this, the maximum quantization error is equal to half of one bit. That leads to a dynamic range of:

$$
\text{dynamic range} = \frac{2^N-1}{0.5} = 2(2^N-1)
$$

The extra factor of 2 gives you an approximate increase of 1 bit in "dynamic range" when measured this way.  I assume that's what is meant by an $N$-bit digitized signal providing $N+1$ bits of dynamic range.