The vanishing (?) cohomology of the Milnor fiber Setup. Say we have a germ of a holomorphic function $f:(\mathbb C^{n+1},0)\to (\mathbb C,0)$ with a critical point at the origin. There is an $\epsilon>0$ small enough so that $f$ becomes a fibration when restricted to $$g=f/|f|:S_\epsilon^{2n+1}\setminus f^{-1}(0)\to S^1.$$
The fibres of $g$ are diffeomorphic to one another, and each of them is called a Milnor fibre of $f$ at $0\in\mathbb C^{n+1}$. 
In the Introduction of these notes on vanishing cycles, the author states:

"The cohomology of the Milnor fibre is known to be nonzero in degrees between $0$ and $n$ at most [$\dots$]"

In the next paragraph, the author states that the Milnor fiber $F$ of an isolated singular point has the same homotopy type of a bouquet of $\mu(f)$ spheres, where $\mu(f)$ is the Milnor number of $f$, namely the dimension of the vector space $\mathbb C[[x_0,\dots,x_n]]/(\textrm{Jac}(f))$. In particular, this implies that $H^n(F,\mathbb C)=\mathbb C^{\mu(f)}$, but also that $$H^i(F,\mathbb C)=0,\,\,\,\textrm{if }0<i<n.$$ 

Question 1. Does not this seem to contradict the previous quoted
  assertion? 
Question 2. Am I right in thinking that the cohomology of the Milnor fiber at a regular point of $f^{-1}(0)$ is in fact zero in all
  degrees?

Thank you in advance.

To answer your first question, it is not an error in the mathematics, but simply a poorly written sentence. The Milnor fiber is a smooth manifold of (real) dimension $2n$ but has the homotopy type of a finite CW complex of dimension n, hence the precise statement is that 'the cohomology of the Milnor fiber vanishes above degree $n$.' In fact, as you remark, in the case of an isolated singularity, the fiber is $(n-1)$-connected.
To answer your second question, the Milnor fiber at a regular point is diffeomorphic to $\mathbf{R}^{2n},$ so the answer to your question is yes, if you use reduced cohomology.
