what is the probability of drawing each element at least once from set $\{1,2,3,\dots, n\}$ , drawing with replacements at most $k$ times ($k>n$) what is the probability of drawing each element at least once from set $\{1,2,3,\dots, n\}$, drawing with replacements at most $k_0$ times ($k_0>n$)
my take:
$$\left(\prod_{i=1}^{n} \frac{i}{n}\right)\left(\sum_{j=1}^{n}\left(\frac{j}{n}\sum_{k=1}^j\left(\frac{k}{n}\sum_{l=1}^k\frac{l}{n}\dots [k_0-n \text{ times this construction}]\right)\right)\right)$$
and if this is correct I don't see a way to simplify this and I would ask for help simplifying this expression
how I derived this: first part is probability of obtaining all elements, and then you need to account for "failed" rounds.
i.e. for rolling dice once a day for a week:
\begin{align}
P(\text{rolling each side of dice in 7 throws}) 
&=\left(\prod_{i=1}^{6}\frac{i}{6}\right)\sum_{i=1}^{6}\frac{i}{6}
\end{align}

best solution I have so far is function that have complexity $O(n*k_0)$

from math import prod

def coupon_collector_in_k_draws(n, k):
    denominator_count = n
    res = prod(range(1, n+1))
    n_sq = n**5
    while res > n_sq:
        res /= n
        denominator_count -= 1
    k1 = k-n
    tab = [1] * (k1+1)
    den_count = [0] * (k1+1)
    for nn in range(2, n+1):
        for k in range(k1)[::-1]:
            tab[k] += nn*tab[k+1]/(n**(den_count[k] - den_count[k+1]))
            while tab[k] > n_sq and den_count[k] < k1:
                tab[k] /= n
                den_count[k] += 1
    rs = tab[0]
    for j in range(k1 - den_count[0]):
        rs /= n
    res *= rs
    while denominator_count:
        res /= n
        denominator_count -= 1
    return res

 A: Let me put $k_{\,0}  = m$.
Then your problem is equivalent to that of throwing $m$ labelled balls ( = position in the extraction sequence)
into $n$ bins ( = extraction result), in such a way that no bin is left empty.
It is also equivalent to the number of ways to compose a set of $m$ elements into a list (order counts) of $n$ non-empty subsets
which is given by $n!$ times the Stirling Number of 2nd kind
$$
N(m,n) = n!\left\{ \matrix{  m \cr   n \cr}  \right\}
$$
The total number of ways of throwing the balls is clearly $n^m$, which in fact equals
$$
n^{\,m}  = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)}
 {\left\{ \matrix{  m \cr   k \cr}  \right\}\,n^{\,\underline {\,k\,} } } \;
 = \sum\limits_{0\, \le \,k\,\left( { \le \,n} \right)}
 {k!\left\{ \matrix{  m \cr   k \cr}  \right\}\,
 \left( \matrix{  n \cr   k \cr}  \right)} 
$$
where $x^{\,\underline {\,k\,} }$ represents the Falling Factorial.
Regarding your approach, consider that an alternative definition of the Stirling N. 2nd kind is
$$
\eqalign{
  & \left\{ \matrix{  m \cr   n \cr}  \right\}\quad 
 = {1 \over {n!}}\sum\limits_{\left\{ {\matrix{
   {1\, \le \,k_{\,j} \,\left( { \le \,m} \right)}  \cr 
   {k_{\,1}  + k_{\,2}  + \, \cdots  + k_{\,n} \, = \,m}  \cr 
 } } \right.\;} {\left( \matrix{  m \cr   k_{\,1} ,k_{\,2} , \cdots k_{\,n}  \cr}  \right)}  =   \cr 
  &  = {{m!} \over {n!}}\sum\limits_{\left\{ {\matrix{
   {0\, \le \,l_{\,j} \,\left( { \le \,m - n} \right)}  \cr 
   {l_{\,1}  + l_{\,2}  + \, \cdots  + l_{\,n} \, = \,m - n}  \cr 
 } } \right.\;} {{1 \over {\left( {l_{\,1}  + 1} \right)!\left( {l_{\,2}  + 1} \right)!\, \cdots \left( {l_{\,n}  + 1} \right)!}}}  =   \cr 
  &  = \left( \matrix{  m \cr   n \cr}  \right)\sum\limits_{\left\{ {\matrix{
   {0\, \le \,l_{\,j} \,\left( { \le \,m - n} \right)}  \cr 
   {l_{\,1}  + l_{\,2}  + \, \cdots  + l_{\,n} \, = \,m - n}  \cr 
 } } \right.\;}
 {{1 \over {\left( {l_{\,1}  + 1} \right)\left( {l_{\,2}  + 1} \right)\, \cdots \left( {l_{\,n}  + 1} \right)}}
\left( \matrix{  m - n \cr   l_{\,1} ,l_{\,2} ,\, \cdots ,l_{\,n}  \cr}  \right)}  \cr} 
$$
and that the multinomial can be written as
$$
\eqalign{
  & \left( \matrix{  m \cr   k_{\,1} ,k_{\,2} , \cdots k_{\,n}  \cr}  \right)
\quad \left| {\;k_{\,1}  + k_{\,2}  + \, \cdots  + k_{\,n} \, = \,m} \right.\quad  =   \cr 
  &  = {{m!} \over {k_{\,1} !k_{\,2} ! \cdots k_{\,n} !}}
 = {{m^{\,\underline {\,k_{\,1}  + k_{\,2}  + \, \cdots  + k_{\,n} \,} } } \over {k_{\,1} !k_{\,2} ! \cdots k_{\,n} !}} =   \cr 
  &  = {{m^{\,\underline {\,k_{\,1} \,} } } \over {k_{\,1} !}}
{{\left( {m - k_{\,1} } \right)^{\,\underline {\,k_{\,2} \,} } } \over {k_{\,2} !}}
{{\left( {m - k_{\,1}  - k_{\,2} } \right)^{\,\underline {\,k_{\,3} \,} } } \over {k_{\,3} !}}\, \cdots \;
{{\left( {m - \left( {m - k_{\,n} } \right)} \right)^{\,\underline {\,k_{\,n} \,} } } \over {k_{\,n} !}} \cr} 
$$
