If the number of elements of two finite sets are different, then they have different cardinality. I found the below proof a little weired. Would anyone give me some suggestions please (if it is correct at all)? In particular in the section arguing the range has at least $n$ elements. 

Claim: If $n,m\in\mathbb{N}_{\geq 0}$ and $n>m$, then $\mathbb{N}_n \not\approx \mathbb{N}_m$.

Some Notations: $\mathbb{N}_n=\{e\in\mathbb{N} \mid e\leq n\}$; $\approx$ is the relation of equal cardinality.
Proof: Suppose $n,m\in\mathbb{N}_{\geq 0}$ and $n>m$. Suppose $\mathbb{N}_n \approx \mathbb{N}_m$ for the sake of contradiction. In other words, there exists a bijective mapping $f: \mathbb{N}_n \rightarrow \mathbb{N}_m$. Consider the range of $f$: $f[\mathbb{N}_n]$. Notice that $f(i), f(j)\in f[\mathbb{N}_n]$ and $f(i)\neq f(j)$ for all $i,j\in \mathbb{N}_{n}$. Hence there is at least $n$ element in the range and $f[\mathbb{N}_n]\subseteq \mathbb{N}_m$ since it is the range. Therefore, $\mathbb{N}_m$ has at least $n$ elements, a contradiction. Hence we conclude that $\mathbb{N}_n \not\approx \mathbb{N}_m$ as desired.

I believe you are using the Pigeonhole principle at the end of your proof. I believe it is a valid proof.
