Coupon collector's problem using inclusion-exclusion Coupon collector's problem asks:

Given n coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?

The well-known solution is $E(T)=n \cdot H_n$, where T is the time to collect all n coupons(proof).
I am trying to approach another way, by calculating possible arrangements of coupons using inclusion-exclusion(Stirling's numbers of the second kind) and that one coupon should only be collected at last and other coupons should be collected at least once:
$$P(T=k)=\frac{n!\cdot{k-1\brace n-1}}{n^k}\\
=\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\
E(T)=\sum\limits_{k=n}^{\infty}k\cdot P(T=k)\\
=\sum\limits_{k=n}^{\infty}k\cdot\frac{\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot i^{k-1}}{n^{k-1}}\\
=\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot\sum\limits_{k=n}^{\infty}k\cdot (\frac i n)^{k-1}\\
=\sum\limits_{i=1}^{n-1}(-1)^{n-i-1}\cdot{n-1\choose i}\cdot(\frac i n)^{n-1}\cdot(\frac 1 {1-\frac i n})\cdot(n-1+\frac 1 {1-\frac i n})$$
Calculation of first 170 terms yields same results.
Are two formulas same?
 A: By way of enrichment here is  a proof using Stirling numbers of the
second kind  which encapsulates inclusion-exclusion  in the generating
function of these numbers.

First let  us verify  that we indeed  have a  probability distribution
here. We have for the number $T$ of coupons being $m$ draws that
$$P[T=m] = \frac{1}{n^m} \times 
n\times {m-1\brace n-1} \times (n-1)!.$$
Recall the OGF  of the Stirling numbers of the  second kind which says
that
$${n\brace k} = [z^n] \prod_{q=1}^k \frac{z}{1-qz}.$$
This gives for the sum of the probabilities
$$\sum_{m\ge 1} P[T=m]
= (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}} {m-1\brace n-1}
\\ = (n-1)! \sum_{m\ge 1} \frac{1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = (n-1)! \prod_{q=1}^{n-1} \frac{1/n}{1-q/n}
= (n-1)! \prod_{q=1}^{n-1} \frac{1}{n-q} = 1.$$
This confirms it being a probability distribution.
We then get for the expectation that
$$\sum_{m\ge 1} m\times P[T=m]
= (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}} {m-1\brace n-1}
\\ = (n-1)! \sum_{m\ge 1} \frac{m}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + (n-1)! \sum_{m\ge 1} \frac{m-1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + (n-1)! \sum_{m\ge 2} \frac{m-1}{n^{m-1}}
[z^{m-1}] \prod_{q=1}^{n-1} \frac{z}{1-qz}
\\ = 1 + \frac{1}{n} (n-1)! \sum_{m\ge 2} \frac{1}{n^{m-2}}
[z^{m-2}] \left(\prod_{q=1}^{n-1}
\frac{z}{1-qz}\right)'
\\ = 1 + \frac{1}{n} (n-1)! 
\left.\left(\prod_{q=1}^{n-1}
\frac{z}{1-qz}\right)'\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)! 
\left. \left(\prod_{q=1}^{n-1} 
\frac{z}{1-qz}
\sum_{p=1}^{n-1} \frac{1-pz}{z} \frac{1}{(1-pz)^2}
\right)\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)! 
\prod_{q=1}^{n-1} \frac{1/n}{1-q/n}
\left. \sum_{p=1}^{n-1} \frac{1}{z} \frac{1}{1-pz}
\right|_{z=1/n}
\\ = 1 + \frac{1}{n} (n-1)! 
\prod_{q=1}^{n-1} \frac{1}{n-q}
\sum_{p=1}^{n-1} \frac{n}{1-p/n}
\\ = 1 + \frac{1}{n}
\sum_{p=1}^{n-1} \frac{n^2}{n-p}
= 1 + n H_{n-1} = n \times H_n.$$
What we have here are in fact two annihilated coefficient extractors (ACE) more of which may be found at this MSE link. Admittedly the EGF better represents inclusion-exclusion than the OGF and could indeed be used here where the initial coefficient extractor would then transform it into the OGF.
