Don't understand the *derivation* of geometrically distributed random variables I don't understand the derivation of geometrically distributed random
variables as done here (only the first $10$ lines - everything until exercise $2$ -  are relevant for me). 
Please bare with me, that I probably need a very formal and thorough explication, since I looked already at different website and books, which used explanations in a similar manner to the link I provided and I didn't understood any of them! (I have my "own" derivation, which makes sense to me, which can be found below)
In the course I took, they weren't rigorously defined, so I more or
less tried to figure out what the professor did and it seems to me
that what is written below is how we derived them. But the derivation
in the link is different! And the central problem seems to be, that
in the link there is a mathematical connection between the random
variables $X_{k}:\left\{ s,f\right\} \rightarrow\mathbb{R},\ \omega\mapsto\omega$
(where our Bernoulli trail sample space is modeled by $\Omega:\left\{ s,f\right\} $,
for " success'' and " failure'' and the whole experiment
is $\Omega^{\mathbb{N}}$ and $k\in\mathbb{N}$), since there they
somehow used that the $X_{k}$ are independent -- which doesn't make
sense at all to me: If $s$ has probability $1>p>0$, then the sets
$\left\{ X_{1}=s\right\} $ and $\left\{ X_{2}=s\right\} $ aren't
independent, because 
$$
P\left(\left\{ X_{1}=s\right\} \cap\left\{ X_{2}=s\right\} \right)=P\left\{ s\right\} =p\neq p^{2}=P\left(\left\{ X_{1}=s\right\} \right)P\left(\left\{ X_{2}=s\right\} \right).
$$
Opposed to that, in my derivation there isn't a mathematical link
between the random variable that has the geometric distribution and
the $X_{k}$.

My derivation: Consider sequence of Bernoulli trials of finite length $n$ (where
the space for each experiment is for example $\Omega:\left\{ s,f\right\} $).
If $p$ is the probability to get success, $p(1-p)^{k-1}$ is the
probability of having the first success in the $k$th trial, where
$n\geqslant k\left(\geqslant0\right)$. Now we want to know what the
probability is to have first success in the $k$th trial, where we
don't have an upper bound $n$.
Since we don't have a way to define on the space of all infinite sequences,
i.e. $\left(\omega_{1},\omega_{2},\ldots\right)$ mit $\omega_{i}\in\Omega$
a suitable probability distribution starting from our $\Omega$ (at
least I don't know how to do it and using advanced measure theoretic
machinery doesn't count, since I'm interested in hoe the geometric
distribution is defined in " undergraduate'' setting). But we
do know that $p(1-p)^{k-1}$ is independent of $n$, so we define
a new $\hat{\Omega}:=\mathbb{N}$ und $\hat{p}\left(k\right):=p\left(1-p\right)^{k}$
we models the fact that we have first success in the $k$th trial.
Then we can define $X:\hat{\Omega}\rightarrow\mathbb{R},\ x\mapsto x$
as the random variable whose distribution is which tells us, when
we get the first succes - but mathematically there is no connection
between this random variable, and the random variables $X_{k}:\left\{ s,f\right\} ^{n}\rightarrow\mathbb{R},\ \left(\omega_{1},\omega_{2},\ldots,\omega_{n}\right)\mapsto\omega_{k}$
or the random variables $X_{k}:\left\{ s,f\right\} \rightarrow\mathbb{R},\ \omega\mapsto\omega$. 
 A: I’ve tried to summarize in this answer some of the discussion in the comments.
The root of your difficulty, I think, is confusion about the appropriate sample spaces. On the one hand you have the Bernoulli random variables $X_k$ for $k\in\Bbb Z^+$; each of them is a function on the sample space $\Omega$ taking values in $\{0,1\}$. We define $P(\{s\})=p$ and $P(\{f\})=1-p$, from which we have automatically that $$P_{X_k}(\{1\})=P(X_k=1)=P(\{s\})=p$$ and $$P_{X_k}(\{0\})=P(X_k=0)=P(\{f\})=1-p\;.$$
On the other hand you have the geometric random variable $N$, defined as the least $k\in\Bbb Z^+$ such that $X_k=1$. The experiment in this case is to perform an infinite sequence $\mathbf X=\langle X_1,X_2,X_3,\dots\rangle$ of independent Bernoulli trials, so the possible outcomes are the infinite sequences that are the points of the Cartesian product $\Omega^{\Bbb Z^+}$. 
Note that independence of the Bernoulli trials is part of the definition of a geometric random variable. Thus, even without access to any measure theory we can naïvely argue that
$$\begin{align*}
P_N(\{k\})&=P(N=k)\\
&=P(X_1=0~\&~X_2=0~\&~\ldots~\&~X_{k-1}=0~\&~X_k=1)\\
&\overset{*}=P(X_1=0)\cdot P(X_2=0)\cdot\ldots\cdot P(X_{k-1}=0)\cdot P(X_k=1)\\
&=(1-p)^{k-1}p\;,
\end{align*}$$
using independence of the $X_i$ to justify the starred equality.
However, we can also look at the set of outcomes resulting in $N=k$: it’s
$$\left\{\omega\in\Omega^{\Bbb Z^+}:N(\omega)=k\right\}=\left\{\langle\omega_1,\omega_2,\omega_3,\dots\rangle\in\Omega^{\Bbb Z^+}:\omega_1=\ldots=\omega_{k-1}=f\text{ and }\omega_k=s\right\}\;.$$
If for each $n\in\Bbb Z^+$ we let $\Omega_n=\{s,f\}$, we can write this set as
$$\left\{\omega\in\Omega^{\Bbb Z^+}:N(\omega)=k\right\}=\{f\}^{k-1}\times\{s\}\times\prod_{n>k}\Omega_n\;,$$
a cylinder set in the product $\Omega^{\Bbb Z^+}=\prod_{n\ge 1}\Omega_n$.
Note that the probability that we naïvely assigned to this set actually is the product of the probabilities associated with the individual factors: all but finitely many of those are $1$, so in effect it’s a finite product. This assignment of probabilities to cylinder sets is in fact the starting point for constructing a probability measure on the product, and for this particular setting we need nothing more.
