Introduction:
In this answer, we will prove that any group of order 30 must contain an element of order 30. This result is derived from the fact that the order of an element must divide the order of the group.

Proof:

1. The order of an element:
The order of an element g in a group G is the smallest positive integer n such that g^n = e, where e is the identity element of G. This means that the element g, when multiplied by itself n times, gives the identity element.

2. Lagrange's theorem:
Lagrange's theorem states that the order of any subgroup H of a finite group G divides the order of G. In other words, if G has order n and H is a subgroup of G, then the order of H must divide n.

3. Order of elements in a group of order 30:
Let's consider a group G of order 30. By Lagrange's theorem, the order of any element in G must divide 30. Therefore, the order of each element in G can be one of the divisors of 30, which are 1, 2, 3, 5, 6, 10, 15, and 30.

4. Non-identity elements:
Since the identity element has an order of 1, we need to find a non-identity element with an order of 30. Let's consider an element g in G that is not equal to the identity element.

5. Order of g:
If the order of g is 30, then we have found an element of order 30. Otherwise, we need to consider the order of g^2, g^3, g^5, g^6, g^10, g^15, or g^30.

6. Dividing the group order:
If the order of g^k is equal to 30 for any k, then we have found an element of order 30. Otherwise, we consider the orders of g^2k, g^3k, g^5k, g^6k, g^10k, g^15k, or g^30k for different values of k.

7. Conclusion:
Since the order of each element in G divides 30, and we have considered all possible powers of g, we must eventually find an element with an order of 30. Thus, any group of order 30 has an element of order 30.