One-time pad, zero key and Shannon I'm supposed to prove that OTP without the zero key $k=0^n$ is not perfectly secret anymore. I understand that it's not because an attacker learns something by looking at the plaintext and ciphertext. But I don't know how to prove it formally.

I was thinking about using Shannon. But to do so $|K|$ needs to be equal to $|P|=|C|=|{0,1}^n|$, which is not the case if I define $K={0,1}^n$ \ $0^n$. So my idea was to include the zero key to the key space K, but set the probability for $p(k=0^n)=0$. Doing so would allow me to have $|K|=|P|=|C|$ and to use Shannon. 

PS: I'm aware of the somewhat similar question (One-time pad and zero key), but we didn't have any rule for $|K|&gt;=|P|$ in order to achieve perfect secrecy. Also I am interested in my question about Shannon.
Well, for perfect secrecy, we require that for all message distributions over $\mathcal{M}$, all messages $m\in\mathcal{M}$, and all (possible) ciphertexts $c$ it holds that

$$Pr[M=m\ |\ C = c] = Pr[M=m]$$

In particular, that means, if we can find a single counterexample, i.e. a distribution over $\mathcal{M}$, a message $m\in\mathcal{M}$, and a ciphertexts $c$ such that

$$Pr[M=m\ |\ C = c] \neq Pr[M=m]$$

then, by our definition, the encryption scheme does not offer perfect secrecy.

So, let's take the uniform distribution over $\mathcal{M}$. For this distribution, we obviously have $$Pr[M=c]=1/|\mathcal{M}|.$$
Now, given that $0^n$ is not a valid key, what is the probability  $$Pr[M=c\ |\ C = c]?$$ Are the two probabilities equal?