Degree of irreducible representation of nontrivial finite group is strictly less than order of group

Statement
Suppose $$G$$ is a nontrivial finite group of order $$n$$ and $$k$$ is a field. Then, any irreducible representation of $$G$$ over $$k$$ has degree at most $$n - 1$$.

Equality is attained for infinitely many groups: namely, where $$G$$ is a group of prime order and $$k$$ is $$\mathbb{Q}$$, the field of rational numbers.

Proof
Given: A group $$G$$ of order $$n$$. A nonzero vector space $$V$$, a homomorphism $$\rho:G \to GL(V)$$ such that $$V$$ has no proper nonzero $$G$$-invariant subspace.

To prove: The dimension of $$V$$ is at most $$n - 1$$.

Proof: Let $$0 \ne v \in V$$. Consider the set:

$$A = \{ g \cdot v \}_{g \in G}$$

The vector space spanned by $$A$$ is $$G$$-invariant and nonzero, hence must equal $$V$$. Since it has a spanning set of size $$n$$, it has dimension at most $$n$$. Further, the dimension equals $$n$$ if and only if $$A$$ is linearly independent. However, if $$A$$ is linearly independent, the element:

$$\sum_{g \in G} g \cdot v$$

is $$G$$-invariant and nonzero, hence it spans a nonzero $$G$$-invariant one-dimensional subspace of $$V$$. The subspace is proper since $$|G| > 1$$, contradicting the irreducibility assumption. Thus, $$A$$ cannot be linearly independent, and the dimension of $$V$$ is at most $$n - 1$$.