The formula shown below can be used to calculate the probability of not collecting all $m$ coupons after $k$ trials in a coupon collector's problem (with $m$ coupons). Could you please explain us something about the "$-1$" in the formula below? Where does it come from and what it is the intuition behind it?
$$p = \sum^{m}_{i=1} (-1)^{i+1} {m \choose i} \left(\frac{m-1}{m}\right)^k$$
For what it's worth, I am posting this answer because my research into the Coupon Collector's problem has failed to find this exact problem addressed.
I may be misinterpreting the OP's (i.e. original poster's) intent. I am interpreting the problem to be that you have $m$ coupons, and that you randomly select $k$ coupons, one at a time, with replacement.
I am assuming that $k \geq m.$
Also, unless I am mistaken, the correct formula is
$$p = \sum^{m}_{i=1} (-1)^{i+1} {m \choose i} \left(\frac{m-\color{red}{i}}{m}\right)^k.$$
Inclusion Exclusion is used. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
For any set $E$, with a finite number of elements, let $|E|$ denote the number of elements in the set $E$.
For the specific problem, let $S$ denote the collection of all multisets that represent all possible distributions of the $m$ coupons, where each set contains $k$ elements, and a specific element (i.e. collected coupon) can occur more than once.
For $i \in \{1,2,\cdots,m\}$, let $S_i$ denote the subset of $S$ that contains all multisets that are missing coupon $i$.
Then, the desired computation is
$$\frac{|S_1 \cup S_2 \cup \cdots \cup S_m|}{|S|} ~: ~|S| = m^k.$$
Let $T_1$ denote $~\displaystyle \sum_{1 \leq i_1 \leq m} |S_{i_1}|.$
Let $T_2$ denote $~\displaystyle \sum_{1 \leq i_1 < i_2 \leq m} |S_{i_1} \cap S_{i_2}|.$
That is, $T_2$ represents the sum of $~\displaystyle \binom{m}{2}~$ terms.
Similarly, for $~3 \leq r \leq m,~$ let $~T_r~$ denote the $~\displaystyle \binom{m}{r}~$ terms given by
$~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq m} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|.$
Then, in accordance with Inclusion Exclusion Theory,
$$|S_1 \cup S_2 \cup \cdots \cup S_m| = \sum_{r=1}^m (-1)^{r+1} T_r.$$
Thus, the entire problem has been reduced to showing that
$$T_r = \binom{m}{r}\left(m-r\right)^k. \tag1 $$
To enumerate $T_r$, first consider the specific computation of
$|S_1 \cap S_2 \cap \cdots \cap S_r|.$
There are $(m-r)^k$ ways that $k$ coupons can be collected from the subset of coupons represented by $\{r+1, r+2, \cdots, m\},~$ where the sampling is done with replacement.
Further, by considerations of symmetry, for any ordered $r$-tuple
$(i_1, i_2, \cdots, i_r) ~: ~i_1 < i_2 < \cdots < i_r,$
you have that
$|S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|$
is also equal to $(m-r)^k.$
So, in computing $T_r$, there are $~\displaystyle \binom{m}{r}~$ ways of selecting the $r$ coupons that will be missing.
So, the formula in (1) above is established.
It is important to note that (for example) in the subset of distributions represented by
$S_1 \cap S_2 \cap \cdots \cap S_r$,
while the coupons represented by $\{1,2,\cdots,r\}$ are missing in each such distribution, other coupons may also be missing from the represented distributions.
In fact, it is this consideration that :
For example, consider the situation where the coupons that are represented by $\{1,2\}$ are missing, and no other coupons are missing.
Such distributions will be counted twice. Once in $|S_1|$ and once in $|S_2|$. Then, these distributions will be deducted once because of the term $|S_1 \cap S_2|$, which is part of the computation of $T_2$.
So, the net effect is that such coupons are counted $~(+2 - 1) = 1~$ time.