Let the balls be labelled $1,2,3,..n$ and the boxes be labelled $1,2,3,..,n$.
Now I want to find,
What is the expected value of the minimum value of the label among the boxes which are non-empty
What is the expected number of boxes with exactly one ball in them?
Whatever way I am thinking of it, I am getting complicated summation form of answers and not any exact closed form!
On
Here is an approach using labelled species and exponential generating functions.
For the first problem we have the species $$\sum_{q=1}^n \mathcal{U}^q \times \mathfrak{P}_{\ge 1}(\mathcal{Z}) \times \mathfrak{P}(\mathcal{Z})^{n-q}$$ with $\mathcal{U}$ marking the end of the intial segment of empty bins.
This yields the generating function $$\sum_{q=1}^n u^q (\exp(z)-1) \exp(z)^{n-q} = (\exp(z)-1) \exp(z)^n \sum_{q=1}^n u^q \exp(z)^{-q}.$$ Some algebra produces $$(\exp(z)-1) \exp(z)^{n-1} \times u \frac{(u/\exp(z))^n - 1}{u/\exp(z)-1}.$$ which is $$(\exp(z)-1) \frac{u^{n+1} - u \exp(z)^n}{u-\exp(z)}.$$
Differentiate and put $u=1$ to obtain the EGF of the count $$\left.(\exp(z)-1) \left(\frac{(n+1)u^n - \exp(z)^n}{u-\exp(z)} - \frac{u^{n+1} - u \exp(z)^n}{(u-\exp(z))^2} \right)\right|_{u=1} \\= (\exp(z)-1) \left(\frac{(n+1) - \exp(z)^n}{1-\exp(z)} - \frac{1 - \exp(z)^n}{(1-\exp(z))^2} \right) \\ = \frac{1 - \exp(z)^n}{1-\exp(z)} + \exp(z)^n - (n+1) = -(n+1) + \frac{1 - \exp(z)^{n+1}}{1-\exp(z)}.$$
Performing coefficient extraction we obtain for $n\ge 1$ the answer $$\frac{1}{n^n} n! [z^n] \left(-(n+1) + \frac{1 - \exp(z)^{n+1}}{1-\exp(z)}\right) = \frac{1}{n^n} n! [z^n] \left(-(n+1) + \sum_{q=0}^n \exp(z)^q\right) \\ = \frac{1}{n^n} n! \sum_{q=1}^n \frac{q^n}{n!} = \frac{1}{n^n} \sum_{q=1}^n q^n = 1 + \frac{1}{n^n} \sum_{q=1}^{n-1} q^n.$$
For the second problem the species is $$\mathfrak{S}_{=n} \left(\mathcal{U}\mathcal{Z}+\mathfrak{P}_{\ne 1}(\mathcal{Z})\right).$$ with $\mathcal{U}$ marking singletons.
This gives the generating function $$\left(uz + \exp(z) - z\right)^n.$$
For the expected number of singletons differentiate with respect to $u$ and set $u=1$ to obtain
$$\left. n \left(uz + \exp(z) - z\right)^{n-1} \times z\right|_{u=1} = n z \exp(z)^{n-1}.$$
Finally extract coefficients for the expectation which is $$\frac{1}{n^n} n! [z^n] n z \exp(z)^{n-1} = \frac{1}{n^{n-1}} n! [z^{n-1}] \exp(z)^{n-1} = \frac{1}{n^{n-1}} n! \frac{(n-1)^{n-1}}{(n-1)!} \\= \frac{1}{n^{n-1}} n (n-1)^{n-1} = \frac{(n-1)^{n-1}}{n^{n-2}}.$$
These results match the first answer.
Remark. Since $$\frac{(n-1)^{n-1}}{n^{n-2}} = (n-1)\left(1-\frac{1}{n}\right)^{n-2} = (n-1)\left(1-\frac{1}{n}\right)^n \left(1-\frac{1}{n}\right)^{-2} \\ = \frac{n^2}{n-1} \left(1-\frac{1}{n}\right)^n$$ the second expectation is $$\frac{1}{e} \frac{n^2}{n-1}.$$
A
To expectation of the minimum used label, $K$, we first measure the probability that all the balls being among the top $n-k$ boxes. That is, that the minimum label will be greater than some value $k$.
In the total space each of $n$ balls has a choice of $n$ boxes ($n^n$). In the restricted space each of $n$ balls has a choice of $n-k$ boxes $(n-k)^n$. So then: $$\begin{align} \Pr(K > k) & = \frac{(n-k)^n}{n^n} \\[1ex] \Pr(K > k-1) & = \frac{(n-k+1)^n}{n^n} \\[1ex] \Pr(K=k) & = \Pr(K>k-1)-\Pr(K>k) \\ & = \frac{(n-k+1)^n-(n-k)^n}{n^n} \\[2ex] E(K) & = \sum_{k=1}^{n} k \Pr(K=k) \\ & = \frac 1 {n^n}\sum_{k=1}^n (k(n-k+1)^n-k(n-k)^n) \\ & = \frac 1 {n^n}\left(\sum_{k=0}^{n-1} (k+1)(n-k)^n-\sum_{k=1}^{n} k(n-k)^n\right) \\ & = \frac 1 {n^n}\left( n^n + \sum_{k=1}^{n-1} (n-k)^n\right) \\ & = \frac 1 {n^n}\left( n^n + \sum_{k=1}^{n-1} k^n\right) \\ & = \mathop{1 + n^{-n}\underbrace{\sum_{k=1}^{n-1} k^n}}_{\text{a generalised Harmonic series?}} \end{align}$$
B
Let $B_i$ be the Bernouli indicator that box $i$ contains exactly one ball. We then use the linearity of expectation to find the expected value of $B:=\sum_{i=1}^n B_i$
$$\begin{align} \forall i\in \{1..n\}:\Pr(B_i=1) & = {n\choose 1}(n-1)^{n-1}/n^n \\[2ex] \mathbb{E}[B] & = \mathbb{E}[\sum_{i=1}^n B_i] \\ & = \sum_{i=1}^n \mathbb{E}[B_i] \\ & = n\times (0\times \Pr(B_\ast=0)+1\times \Pr(B_\ast=1)) \\[1ex]\therefore \mathbb{E}[B] & = \frac{(n-1)^{n-1} }{ n^{n-2}} \end{align}$$