CDF and Survival Function of Geometric Distribution

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I am having trouble understanding the intuition behind the CDF and survival probability of a geometric distribution on both $\{0, 1, \ldots \}$ and on $\{1, 2, 3, \ldots \}$.

I know that a geometric starting from $0$ is the number of failures before the first success and the geometric starting at 1 is the number of failures including the first success.

Can someone please explain the intuition behind the CDF and survival functions and explain what the formula is?

I think that the part confusing me is: Why is the following not true? $$p\left(X>n\right)=\left(1-p\right)^{n+1}+\left(1-p\right)^{n+2}+...$$ for a geometric starting at $0$.

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If $X$ is the number of failures before the arrival of first success then the event $\{X>n\}$ is the same as the event that the first $n+1$ trials are not a success, so the probability on that is $(1-p)^{n+1}$.

Then consequently:$$F_X(n)=P(X\leq n)=1-P(X>n)=1-(1-p)^{n+1}$$

Similarly we find: $$F_X(n)=P(X\leq n)=1-P(X>n)=1-(1-p)^{n}$$if $X$ denotes the number of trials need to arrive at the first success.


Concerning your question note that $(1-p)^{n+k}$ is the probability that the first $n+k$ trials are failures.

However these events for $k=1,2,3,\dots$ are not mutually exclusive.

So summation of the terms leads to multiple counting.