Geometric distribution, showing that $P(X>x)=(1-p)^{x}$ from the pmf definition

This may be a silly question but I can't seem to solve this for the life of me!

The question states:

Let the random variable $X$ have a Geometric distribution with parameter $p$ and probability mass function:

$p(x)=p(1-p)^{x-1}$ ; $x=1,2,3,...$

$X$ can represent the number of the trial on which the first success occurs in an infinite sequence of independent Bernoulli trials each with parameter $p$.

So from the definition of the probability mass function, I must show that:

$P(X>x)=(1-p)^{x}$ ; $x=1,2,3,...$

Any advice is appreciated!

Solution 1:

Observe that

$$P(X > x) = P(\hbox{all fails for the first $x$ trials}) = (1 - p)^x.$$

Solution 2:

As Dilip Sarwate suggested, the Geometric Series is most applicable to this Geometric Distribution problem.

$$\begin{align}\mathsf P(X>x) &=\mathbf 1_{x\in\Bbb Z^+}\sum_{k=1}^\infty \mathsf p(x+k)\\[1ex] &=\mathbf 1_{x\in\Bbb Z^+}\sum_{k=1}^\infty p~(1-p)^{x+k-1}\\[0ex] &~~\vdots\end{align}$$