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I'm trying to prove the following but I'm not sure where to even start.

Given is a finite group $G$ with an order $> 1$ and that all elements of the group except for the $1$ have the same order $= k$. Prove that $k$ must then be prime.

Any help is appreciated.

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    Suppose that $g \in G$ has order $k$ which is not prime, i.e. you have $k = nm$ with $n,m > 1$. Then.... – 2017-02-13

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To show this first show the following:

Let $g$ be an element of order $k$ and let $d \mid k$. Then the order of $g^d$ is $k/d$.

Having established this you should be able to conclude easily.

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    There must be some property of the order of an element that I am missing... Im looking at my script but Im not sure. – 2017-02-13
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    The order of an element is the smallest positive integer $k$ such that $g^k = 1$. So you need to show that $(g^d)^{k/d}= 1$ and $(g^d)^r \neq 1$ for $1 \le r < k/d$. To do so you will want to use that $g^k = 1$ and $g^s \neq 1$ for $1 \le s < k$. – 2017-02-13
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    I think I have it now, can you let me know if I'm correct? I assumed that $ord(g) = k$ is not a prime number, then there must exist a $d$ such that $d \mid k$. Since we assumed that all elements of the group must have the same order, then $g^d$ is not equal to $1$. And there is a rule that says $ord(g^d) = \frac{ord(g)}{GCD(d, ord(g))}$ and then I get that $ord(g^d) = \frac{k}{d}$ which is not k, which is a contradiction to the original assumption that all elements are of the same order. I think this should be correct, right? – 2017-02-13
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    Yes, that's correct. Just make sure to state that $d$ is a proper divisor of $k$ that is it is neither $1$ nor $k$. It is clear you meant this, but it could be considered important to state it explicitly. (You do not strictly need the $g^d$ not $1$ part.) – 2017-02-13