Find the probability that the white ball labelled $1$ is drawn before all the black balls.

The question is rather badly posed. "The first white ball" sounds as if it refers to the first white ball drawn, but apparently it's intended to refer to the white ball with the number $1$.

There's indeed a much simpler way to show that the probability of drawing the white ball with the number $1$ before all black balls is $\frac1{21}$. Consider the $21$ balls comprising the white ball with the number $1$ and the $20$ black balls. All orders of these $21$ balls are equally likely; in particular, the white ball with the number $1$ is equally likely to be in any of the $21$ positions in this order. Thus the probability for each of the positions, including the first position, is $\frac1{21}$.

On how to evaluate your sum, see the hockey-stick identity.

To evaluate $\sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}$:

\begin{align} \sum_{k=1}^{10}\frac{9!}{30!}\frac{(30-k)!}{(10-k)!}&=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\frac{(30-k)!}{20!(10-k)!} \\ &=\frac{9!\cdot 20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20} \\ &=\frac{9!\cdot 20!}{30!}\left[\binom{20}{20}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\ &=\frac{9!\cdot 20!}{30!}\left[\binom{21}{21}+\binom{21}{20}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\ &=\frac{9!\cdot 20!}{30!}\left[\binom{22}{21}+\binom{22}{20}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\ &=\frac{9!\cdot 20!}{30!}\left[\binom{23}{21}+\binom{23}{20}+\ldots+\binom{29}{20}\right] \\ &=\cdots \\ &=\frac{9!\cdot 20!}{30!}\binom{30}{21} \\ &=\frac{9!\cdot 20!}{30!}\frac{30!}{21!\cdot 9!} \\ &=\frac{1}{21} \end{align}

When handling the sum, the relations $\binom{n}{n}=\binom{n+1}{n+1}$ and $\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$ are used.

Your expression $\sum_{k=1}^{10}\frac{9!(30-k)!}{30!(10-k)!}$ can be rewritten as $\frac{9!20!}{30!}\sum_{k=1}^{10}\binom{30-k}{20}$. Now $\binom{30-k}{20}$ is the number of ways to choose $21$ numbers from $\{1,\ldots,30\}$ with the first being $k$ (because the remaining $20$ numbers must be chosen from the $30-k$ numbers after $k$), so summing this over all possible values of the first number chosen (which must be in the range 1-10) gives all ways to choose $21$ numbers from $30$. Thus $\sum_{k=1}^{10}\binom{30-k}{20}=\binom{30}{21}=\frac{30!}{9!21!}$, and so the whole expression becomes $\frac{9!20!}{30!}\times\frac{30!}{9!21!}=\frac{1}{21}$.