Let $I$ be a finite set, $k \in \mathbb{N}$ and $2 \leq k \leq \lvert I \rvert$. Then, the number of subsets of $I$ that have cardinality at most $k$ is bounded above by $\lvert I \rvert^k$, i.e.
\begin{equation*} \sum_{S \subseteq I, \vert S \vert \leq k}1 \leq \lvert I \rvert^k. \end{equation*}
My proof considers two cases:
If $k = 2$, \begin{equation*} \sum_{S \subseteq I, \vert S \vert \leq k} = 1 + \vert I \vert + \frac{\vert I \vert(\vert I \vert-1)}{2} \leq \vert I \vert^k. \end{equation*} If $k \geq 3$, \begin{equation*} \sum_{S \subseteq I, \vert S \vert \leq k} = \sum_{i = 0}^k {\vert I \vert \choose i} < 1 + \vert I \vert + \vert I \vert^{k-1}(\vert I \vert - 1) < \vert I \vert^k. \end{equation*}
I think this proof is not very elegant. Is there a better way of doing it?
Thanks in advance!