Prove that $T_{1}, T_{2}-T_{1}, \dots$ are independent Let $X_{n}$ be a sequence of independent random variables following a Bernoulli distribution of parameter $p$.
We define $T_{0} = 0$ and
 $\forall n \geq 1,$ $$T_{n} = \inf \lbrace k > T_{n-1} ; X_{k} = 1 \rbrace$$ or $T_{n} = \infty$ if the infimum is not finite.
I'm trying to prove that $T_{1}, T_{2}-T_{1}, \dots, T_{n}-T_{n-1}$ are independent. 
Let $(k_{1}, \dots, k_{n}) \in \mathbb{N}^{n}$ ; I have shown that $$P(T_{1} = k_{1}, T_{2}-T_{1} = k_{2}, \dots, T_{n}-T_{n-1} = k_{n}) = (1-p)^{k_{1}-1}p \cdots (1-p)^{k_{n}-1}p$$
Then, $\forall i \geq 1$, 
$$P(T_{i}-T_{i-1} = k_{i}) = \sum\limits_{(k_{1}, \dots, k_{i-1})}P(T_{1} = k_{1}, \dots, T_{i}-T_{i-1} = k_{i})$$
But then I don't know what to do...
 A: $$
\Pr(T_1 = k_1, T_2-T_1 = k_2, \dots, T_n-T_{n-1} = k_n) = (1-p)^{k_1-1}p \cdots (1-p)^{k_n-1}p
$$
First, see if you can show that it is enough to show that for $i=1,\ldots,n$
$$
\Pr(T_i = k_i ) = (1-p)^{k_i}\cdot\text{some constant},
$$
where "constant" means anything that does not depend on $k_i$.
Then you have
\begin{align}
\Pr(\,\overbrace{T_1 = k_1}^{i=1}\,) & = \sum_{\underbrace{k_2,\ldots,k_n}_{2\text{ through }n}} \Pr(\,\overbrace{T_1=k_1}^{i=1}\ \&\ \underbrace{T_2=k_2\ \&\ \cdots\ \&\ T_n=k_n}_{2\text{ through }n}) \\[10pt]
& = \sum_{\underbrace{k_2,\ldots,k_n}_{2\text{ through }n}} \overbrace{(1-p)^{k_1-1}p}^{i=1} \cdot \underbrace{(1-p)^{k_2-1}p \cdots (1-p)^{k_n-1}p}_{2\text{ through }n} \\[10pt]
& = \overbrace{(1-p)^{k_1-1} p}^{i=1} \sum_{\underbrace{k_2,\ldots,k_n}_{2\text{ through }n}} \underbrace{(1-p)^{k_2-1}p \cdots (1-p)^{k_n-1}p}_{2\text{ through }n} \\ & {}\quad (\text{This last step is possible because the factor that} \\
& {}\qquad \text{was pulled out does not depend on $k_2,\ldots,k_n$.}) \\[10pt]
& = \Big((1-p)^{k_1-1} p \times (\text{something that does not depend on $k_1$}) \Big). \end{align}
It's not hard to show that the constant described as "something that does not depend on $k_1$" is $1$.
By the same argument, you can show that
\begin{align}
\Pr(T_2-T_1 = k_2) & = (1-p)^{k_2-1} p \\
\Pr(T_3-T_2 = k_3) & = (1-p)^{k_3-1} p \\
& \vdots \\ & \vdots
\end{align}
So you get
\begin{align}
& \Pr(T_1 = k_1\ \&\ T_2-T_1 = k_2\ \&\ T_3-T_2 = k_3\ \&\ \cdots\ \&\ T_n-T_{n-1}=k_n) \\[10pt]
= {} & \Pr(T_1 = k_1) \cdot \Pr(T_2-T_1 = k_2) \cdot \Pr(T_3-T_2 = k_3) \cdots \Pr(T_n-T_{n-1}=k_n)
\end{align}
and consequently you have independence.
