sequences and series - First Success distribution PMF sum problem

I have the following problem:

Each toss of a coin results in a head with probability $p$. The coin is tossed until the first head appears. Let $X$ be the total number of tosses (including the count of the first head appears).

What is $P(X < m)$ for $m = 1,2, \dots$?

It is important to clarify that, in this context, the variable $X$ is said to have the first success distribution (which is defined to include the toss of the first success), as opposed to the Geometric distribution (which is defined to exclude the toss of the first success). Therefore, if have that $X \sim \text{FS}(p)$ and $Y \sim \text{Geom}(p)$, then $P(X = Y + 1 = k) = P(Y = k - 1) = (1 - p)^{k - 1} p$ for all tosses $k = 1, 2, \dots$, since the PMF for Geometric random variables is $(1 - p)^k p$.

I have that

$$P(X < m = 1) = 0$$

and

$$\begin{align} P(X < m = 2, 3, \dots) &= \sum_{k = 1}^{m - 1} (1 - p)^{k - 1}p \\ &= p + (1 - p)p + \dots + (1 - p)^{m - 2} \\ &= p(1 + (1 - p) + \dots + (1 - p)^{m - 2})\end{align}$$

The solution has that

$$\begin{align} P(X < m = 2, 3, \dots) &= \sum_{k = 1}^{m - 1} (1 - p)^{k - 1}p \\ &= p + (1 - p)p + \dots + (1 - p)^{m - 2} \\ &= p(1 + (1 - p) + \dots + (1 - p)^{m - 2}) \\ &= p\dfrac{1 − (1 − p)^{m−1}}{1 - (1 - p)} \\ &= 1 − (1 − p)^{m−1} \end{align}$$

How did the author get that $p(1 + (1 - p) + \dots + (1 - p)^{m - 2}) = p\dfrac{1 − (1 − p)^{m−1}}{1 - (1 - p)} = 1 − (1 − p)^{m−1}$?

I would greatly appreciate it if people could please take the time to clarify this.

Answer

This is a finite geometric series, and finite geometric series can be easily summed:
$$a+ar+\dots+ar^n=a\frac{1-r^{n+1}}{1-r}$$

Thus
$$1+(1-p)+\dots+(1-p)^{m-2}=\frac{1-(1-p)^{m-1}}{1-(1-p)}$$
and the result follows.