Let $M$ be a finitely generated left $\Lambda V$-module with finite projective dimension $n\geq0$.
Since $\Lambda V$ is a finite dimensional algebra, there exists a minimal projective resolution by finitely generated projectives $0\to P_n\to P_{n-1}\to\cdots\to P_1\to P_0\to M\to 0$ Suppose $n>0$. The module $P_n$ is projective. Since $\Lambda V$ is local, every f.g. projective is free, so in particular $P_n$ is free. Since $\Lambda V$ is an injective module over itself, $P_n$ —being a finite direct sum of copies of $\Lambda V$—is itself injective. This implies that the injection $P_n\to P_{n-1}$ is split, and this is absurd because the resolution was supposed to be minimal.
It follows that necessarily $n=0$.
We conclude in this way that every finitely generated module over $\Lambda V$ whose projective dimension is finite is in fact projective —we say that $\Lambda V$ has «finitistic global dimension equal to zero».
Now, if $\Lambda V$ were of finite global dimension, then we would have that it is in fact of global dimension zero. But it isn't: for example, if we let $S$ be the simple module $\Lambda V/\operatorname{rad}\Lambda V$, there is a surjective map $\Lambda V\to S$ which is not split, as you can easily check, so that $S$ is not projective.
N.B. I used above the fact that $\Lambda V$ is an injective module over itself. We say that it is therefore a self-injective algebra. Such an algebra, when finite dimensional, is always either semisimple or of infinite global dimension, by the same reasoning as above.