In the u.g. book on combinatorics, by David Mazur, there is question #13 on proof in sec. 1.3.
Note: The book, use $[n]$ to denote the set of the first $n$ positive integers.
This exercise outlines a bijective proof of the formula $\left(\!\!\!\binom{n}{k}\!\!\!\right) = \binom{k+n-1}{k}$ from Section 1.1. Let $A $ be the set of $k$-multisets taken from $[n]$ and let $B$ be the set of $k$-subsets of $[k+n-1]$. Assume that the $k$-multiset $\{a_1,a_2,\ldots,a_k\}$ is written in non-decreasing order: $a_1 \le a_2 \le\ldots\le a_k$. Define $f: A \to B$ by $$ f({a_1, a_2, ..., a_k}) = {a_1, a_2+1, a_3+2,\ldots, a_k+k-1}.$$ This function, and proof, is originally due to Euler.
(a) Prove that the outputs of $f$ are indeed $k$-subsets of $[k + n - 1]$. This requires proof since it is not immediately clear from the definition of $f$.
(b) Prove that $f$ is a bijection.
My approach is given below for part (a) & request vetting & help (as in the final para.):
Out of $n$ unique elements, $k$ are chosen; with their size (how many of each type are chosen) being in non-decreasing order.
(a) For the function $f$, the elements are non-decreasing (weakly ordered) in domain (set $A$). However in range, the elements are in increasing order by addition of $j-1$ for $j$-th element. In set $B$, the $j$-th element in $ f({a_1, a_2, ..., a_k})$ is $a_j +j -1$.
The lower limit of values in set $B$ (i.e. the smallest value chosen as number of elements for any of the $k$ elements; let, in ordering be $a_1$) is $\ge 1$.
While the upper limit is derived below:
Since $\forall j \in [k]$:
(i) $(1\le a_1 \le a_j)$ & $(0\le j-1) \implies 1\le a_j +j - 1$.
(ii) $(a_j \le a_k \le n)$ & $(j-1 \le k-1)$
From (i) & (ii), get:
$\forall j \in [k],\,\, 1\le a_j +j - 1 \le k +n -1$, with $(k+n-1)$ being the upper limit.
So, all the different elements of $f({a_1, a_2, ..., a_k})$ belong to $[n+k-1]$, i.e. the set of first $n+k-1$ positive integers.
Next, need show that all the mappings are unique, as need show that the outputs of $f$ are unique. This can be shown by proving that the elements in range are in increasing order, i.e. none of the range elements is repeated.
This can be easily shown by taking the two components of the mapping $f(j)=a_j+j-1$, i.e. $a_j$ & $j-1$; out of which $a_j$ can repeat, but the second part is unique.
However, how to show that it is not possible for the quantity $a_j+j-1$ to not repeat is not clear. Given below is an example to elaborate the confusing part:
Let $n=6, k=4, a_1 = 1, a_2 = 2, a_3 =3, a_4=3$ by which I mean that the elements of domain that are chosen are four with their corresponding no. of elements being $1,2,3,3$. I want to know how it is possible to ensure with the mapping $f$ yielding unique values in range. Say, $i=2, j=4, a_i = 2, a_j = 3, i-1 = 1, j-1 = 3$, and it is clear that $a_i+i-1 = 3, a_j+j-1= 6$; but how to theoretically prove the uniqueness of $f(a_k)$ for all $k$ is not known.
For part $(a)$, for $i < j$, we have $a_i \le a_j$ and $i-1 < j-1 $,
Hence $$a_i +i-1 < a_j + j-1.$$
In general, if $a \le b$ and $c < d$, then we have $a+c < b+d$.
To see this note that
\begin{align} b+d - (a+c) &= (d-c)+(b-a) \end{align}
$d-c$ is positive and $b-a$ is nonnegative. Hence the sum is positive.