If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues 
Let $T:V\to V$ be a linear operator. If $\dim V=v$ and $\dim(\ker T)=n$, prove that $T$ has at most $v-n+1$ distinct eigenvalues.

I have been working on this proof for a few days and I am not sure what direction to really go with it? I feel like starting with the rank nullity theorem is correct and relating that to the sum of eigenspaces may be my next move. Though I cant think of how to bring these two ideas together to create a fluid proof? Thank you for your help...

I will assume that $V$ is a vector space over some field $k$ because you do not mention the ground field. 
Suppose $T$ has the set of eigenvalues $\lambda_0$, $\lambda_1$, ..., $\lambda_m$. Since distinct eigenspaces are in a direct sum, write $E_i$ for the eigenspace corresponding to $\lambda_i$. Then
$$
v = \dim(V) \ge \dim E_0 + \dim E_1 + \dots + \dim E_m \ge n+m.
$$
where $m$ is the number of non-zero eigenvalues. (The first inequality is because the direct sum of the eigenspaces is a subspace of $V$, and the second one is because the eigenspaces have dimension at least $1$). This means that $m \le v-n$, but the number of eigenvalues is at most $m+1$, hence if $0$ is an eigenvalue we have $m + 1\le v-n+1$ and if not, we still have $m \le v-n \le v-n+1$. 
Hope that helps,
