Show that a prime ring is primitive $R$ is a prime ring in which the zero ideal is the intersection of finitely many maximal left-ideals. 

Show that $R$ is primitive.

It was suggested to me that the best method to use is proof by contradiction. 

My proof has run like this so far:

Suppose $R$ is not primitive; then 

$\exists A_{i} \triangleleft R$ non-zero with $A_{i} \subseteq M_{i}$, 

where $M_{i}\cap  M_{2} \cap ... M_{n} =  &lt;0_{R}&gt;  $.

Then, $A_{1} A_{2} ...A_{n} \subseteq A_{1}\cap A_{2}\cap ...A_{n}$, and 
$A_{1}\cap A_{2}\cap ...A_{n} \subseteq M_{i}\cap  M_{2} \cap ... M_{n} =&lt;0_{R}&gt;$.

But if R is prime, then $A_{1}A_{2}...A_{n} \neq &lt;0_{R}&gt;$.

Is this contradiction valid?
Yes, that is basically exactly what I would have said: good job.
Yes, that is basically exactly what I would have said: good job.