I am a bit confused as to how the $k$ ball limit is imposed here. In the solution below, I assumed that there were an unlimited supply of balls.
After the $(j{-}1)^{st}$ bin gets a ball, the expected time before the $j^{th}$ bin has a ball is $\frac{n}{n-j+1}$, thus the expected time to fill all $n$ bins would be $ n\sum_{j=1}^n\frac{1}{n-j+1}=n\sum_{j=1}^n\frac{1}{j} $
The argument above uses the Linearity of Expectation. The part that seems a bit magical upon first sight is that $\mathrm{E}[X+Y\;]=\mathrm{E}[X]+\mathrm{E}[Y\;]$ even when $X$ and $Y$ are not independent.
Below is an argument that computes all the probabilities and yields the same answer.
The Generalized Principle of Inclusion-Exclusion says
Let $\{S_i\}$ be a finite collection of sets from a finite universe. Let $N_j$ be the the sum of the sizes of all intersections of $j$ of the $S_i$; thus, $N_0$ is the size of the universe. Then, the number of elements in exactly $k$ of the $S_i$ is $ \sum_{j\ge k}(-1)^{j-k}\binom{j}{k}N_j $ Suppose we've thrown $k$ balls. If we let $S_i$ be the arrangements where bin $i$ is empty, then we have $ N_j=\binom{n}{j}(n-j)^k $ that is, $\binom{n}{j}$ ways to choose the $j$ empty bins and $(n-j)^k$ ways to place the $k$ balls into the remaining $n-j$ bins.
Since there are $n^k$ ways to throw $k$ balls into $n$ bins, the probability of getting exactly $1$ bin empty after throwing $k$ balls is $ \frac{1}{n^k}\sum_{j\ge1}(-1)^{j-1}\binom{j}{1}\binom{n}{j}(n-j)^k $ The probability of getting the last bin with the next ball is $\frac{1}{n}$ and that would be after $k+1$ balls. Thus, the expected number of balls would be $ \begin{align} &\frac{1}{n}\sum_{k\ge0}\sum_{j\ge1}(-1)^{j-1}\binom{j}{1}\binom{n}{j}(k+1)(1-j/n)^k\\ &=\frac{1}{n}\sum_{j\ge1}(-1)^{j-1}\binom{j}{1}\binom{n}{j}\left(\frac{n}{j}\right)^2\\ &=\sum_{j\ge1}(-1)^{j-1}\binom{n}{j}\frac{n}{j}\\ &=\sum_{j\ge1}(-1)^{j-1}\sum_{k=0}^n\binom{n-k-1}{j-1}\binom{k}{0}\frac{n}{j}\\ &=\sum_{j\ge1}(-1)^{j-1}\sum_{k=0}^n\binom{n-k}{j}\frac{n}{n-k}\\ &=\sum_{k=0}^{n-1}\frac{n}{n-k}\\ &=n\sum_{k=1}^n\frac{1}{k} \end{align} $