prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors prove that: there are exactly ${n-1 \choose k-1}$products that consist of $n-k$ factors,so that all these factors are elements of $[k]$. Repetition of factors is allowed,
that is my attempt  :
for explain my attempt let's take $n=7$ and $k=4$
so let's find the number of all product that have $7-4=3$ factors , so that all these factors are element of $[4]$
our concern is just finding the numbers of all these  products , now we know $[4]=\{1,2,3,4\}$, but let's denote it by $\{e_1,e_2,e_3,e_4\}$,
note that $(e_k=k )$ $,k \in \mathbb{Z^+}$
we want to chose $3$ factors from $\{e_1,e_2,e_3,e_4\}$,(repetition of factors is allowed)
all ways to do that equal the number of all solutions of this equation $e_1+e_2+e_3+e_4=3$
for more explain let's take this solution $(2,1,0,0)$ that mean we choose this product $ 1.1.2$,
and if we take  $(1,1,0,1)$ that mean we choose this product $1.2.4$
so the numbers of all products that have 3 factors from $[4]$equal the number  of all solution of the above equation ,and that's equal ${7-4+4-1 \choose 4-1}$
and we can do the same thing with any $n$ and $k$
so finally products that have $n-k$ factors,so that all these factors are element of $[k]$   is a $n-k$-element  multisets of the set $\{1,2,3,4,....k\}$,so the number of all this  products is the number of all weak composition of $ n-k$ into $k$ parts and it is ${ n-k+k-1 \choose k-1}={ n-1 \choose k-1}$
does my attempt true?

Your identification of this problem with the number of solutions of $$\sum_1^k e_i=n-k$$ is problematical since you have not taken account of the fact that the same product can occur in different ways.
For example $$2\times4\times6=2\times3\times8.$$
