Introduction:
To prove that any group of order 30 has an element of order 30, we will make use of the fact that the order of any element in a group divides the order of the group. Since the order of the group is 30, we will explore the possible orders of elements in the group to find an element of order 30.

Possible Orders of Elements:
Let G be a group of order 30. We know that the order of any element in G must divide the order of G, which is 30. Therefore, the possible orders of elements in G are 1, 2, 3, 5, 6, 10, 15, and 30.

Elements of Order 1:
Every group has an identity element, denoted by e, which has order 1. Therefore, G contains an element of order 1.

Elements of Order 2:
If G contains an element of order 2, then we are done. Otherwise, every non-identity element in G has order greater than 2.

Elements of Order 3:
If G contains an element of order 3, then we are done. Otherwise, every non-identity element in G has order greater than 3.

Elements of Order 5:
If G contains an element of order 5, then we are done. Otherwise, every non-identity element in G has order greater than 5.

Elements of Order 6:
If G contains an element of order 6, then we are done. Otherwise, every non-identity element in G has order greater than 6.

Elements of Order 10:
If G contains an element of order 10, then we are done. Otherwise, every non-identity element in G has order greater than 10.

Elements of Order 15:
If G contains an element of order 15, then we are done. Otherwise, every non-identity element in G has order greater than 15.

Elements of Order 30:
Since G is a group of order 30, it must contain at least one element of order 30. This is because if there were no elements of order 30, the sum of the orders of the elements in G would be strictly less than 30, which is a contradiction.

Therefore, we have shown that any group of order 30 must contain an element of order 30.