My Professor used the following ;
Let n be an integer and p prime, then every element of $\mathbb{Z}/ \ p \mathbb{Z} $ has a unique $n^{th}$ root in $\mathbb{Z}/ \ p \mathbb{Z} $ iff $gcd(n,p-1)=1$
But I don't understand how we can know this is true?
Can anyone help to explain this
Also can anyone tell me if my definition is correct? a has a $n^{th}$ root means there exists an x such that $a^{n}=x$
First of all, you have the wrong definition of nth root. An element $a$ has an nth root if there exists some $x$ such that $x^n = a$.
Now, we may rephrase the theorem as the group homomorphism $f:\mathbb{F}_p^* \to \mathbb{F}_p^*$, $f(X)=x^n$ is injective(and hence bijective) iff $\gcd(n,p-1)=1$.
To see that this is true, look at the $\ker(f)$. If $x\in \ker(f)$ the $x^n = 1$ and the order of $x$ must divide n. But the order of x must also divide $p-1$ by Lagrange's theorem, so $\ker(f)=1$ iff $\gcd(n,p-1)=1$.
Note that @Noah's answer does not use the field structure of $\mathbf F_p$, only the group structure of $\mathbf F_p^*$. In this vein, we could start by noticing that a finite subgroup $G$, say of order $g$, of the multiplicative group $K^*$ of a field $K$ is classically known to be cyclic. Then your question applied to $G$ in place of $\mathbf F_p^*$ would read: if $x$ is a generator of $G$, for which $n$ is $x^n$ also a generator of $G$ ? The answer is of course: iff $(n, g)=1$ (Chinese remainder theorem) .