Show that cardinality of a finite set is unique

Question

Let be $M$ a finite set, $n\in\mathbb{N}$ and $f:M\to\{1,\dots,n\}$ a bijection. Then, $n$ is called the cardinality of $M$, denoted by $|M|=n$.

Show that if there are two bijections $f,g$ such that $f:M\to\{1,\dots,n\}$ and $g:M\to\{1,\dots,m\}$, then $m=n$.

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My approach:

I will prove the statement by induction on $n$:

Base case $n=1$:

We know that $\left(g^{-1}\circ f^{-1}\right):\{1\}\to\{1,\dots,m\}$ must be a bijection. So if $m>1$ it would be a contradiction. Hence, $n=1=m$.

Induction hypothesis with $n\in\mathbb{N}$:

If $f,g$ are two bijections with $f:M\to\{1,\dots,n\}$ and $g:M\to\{1,\dots,m\}$, then $m=n$.

Induction step $n\to n+1$:

We consider two bijections $f,g$ where $f:M\to\{1,\dots,n+1\}$ and $g:M\to\{1,\dots, m\}$. Then, \begin{align*} &f_1:M\setminus\{f^{-1}(n+1)\}\to\{1,\dots,n\},~f_1(k):=f(k)\text{ and}\\ &g_1:M\setminus\{g^{-1}(m)\}\to\{1,\dots,m-1\},~g_1(k):=g(k) \end{align*} must be two bijections as well. Otherwise we get a contradiction concerning the bijectivity of $f$ and $g$. We can apply the induction hypothesis which yields $n=m-1$ and finally $n+1=m$. So the induction step is complete and the statement holds for all $n\in\mathbb{N}$.

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Is this correct?

PS: I know that there already exist some questions on this issue (e.g. Subset of a finite set is finite) but I would like to get some feedback on my approach.