Condition in Shamir Secret Sharing Scheme
I am not sure if this will completely answer the question, but I have a couple observations that might help.

First, note that for the random polynomial $q$, $q(0) = a_0$ is the key -- this is from the way that polynomial interpolation works.  So it is important that the value $q(0)$ is never sent to a participant; $q(i)$ is sent for $0 &lt; i &lt; n$.  This is also why it is important that the prime $p &gt; k$ (where $k$ is the number of participants) -- it is to ensure that I can obtain $k$ "shares" without revealing the key.

Second, the polynomial $q$ is not just random, but also secret.  The fact that the degree of the polynomial $q$ is unknown matters.  When $n-1$ participants come together, they can do anything they want -- they are super clever.  They claim that the key is $w$.  However, there is a unique polynomial $g$ of degree $n-1$ such that 
  $$g(i) = q(i)$$
and 
  $$g(0) = w$$
However, this would be true for any $w$.  Since we know that the secret is given by a polynomial of degree less than $n$, and from the data that the pool of $n-1$ participants have, any value of key supposed is as likely ... no value of key can be ruled out.
   **Note however, that this is not true if the participants know the degree of the polynomial.