Show that there is no integer n with $\phi(n)$ = 14 I did the following proof and I was wondering if its valid. It feels wrong because I didn't actually test the case when purportedly $n$ is not prime, but please feel free to correct me.

Assume there exists $n$ such that $\phi(n) = 14$. Assume $n$ is prime. Then $\phi(n) = n-1$. Then here, n-1 = 14, so n = 15. 
We know since Euler's totient function is multiplicative that $\phi(xy)$ = $\phi(x)\phi(y)$, so $\phi(15) = \phi(3)\phi(5)$, but alas $\phi(3)\phi(5) = 2\cdot4 = 8 \ne 14$. If $n$ is not prime a similar argument follows since we know then that $n$ must be composed of prime numbers by the prime factorization theorem.
If a prime $p\mid n$, then $p-1\mid\phi(n)$. If $\phi(n)=14$, then, since the divisors of $14$ are $\{1,2,7,14\}$, $p\in\{2,3\}$. Thus, $n=2^a3^b$ and $\phi(n)$ is of the form $2^j3^k$. However, there is no factor of $7$ in $2^j3^k$.
This does not hold. What kind of similar argument are you then talking about?

You can use the fact that $\phi$ is multiplicative. Assume $n = p_1^{a_1}p_2^{a_2}...p_t^{a_t}$, then $\phi(p_1^{a_1}p_2^{a_2}...p_t^{a_t})$ $\phi(p_1^{a_1})\phi(p_2^{a_2})...\phi(p_t^{a_t}) = 2 \cdot 7 = 1 \cdot 14$

Use this to arrive at a contradiction.