Binomial Distribution and Marginal Distribution Consider $X_{1} = x$ and $n$ a natural number and $p$ between $0$ and $1$
$X_{1}  \sim B(n, p) =  \binom{n}{x} p^{x} q^{n-x}$ 
$X_{2}$ | $ X_{1} \sim B(X_{1}, p)$
...
$X_{k}$ | $ X_{k-1} \sim B(X_{k-1}, p)$
I am looking for the Marginal Distribution of $X_{k}$. How can I represent  $X_{k}$?
 A: Let's simply look at $X_2$ to start, then
$$\mathbb{P}(X_2=x)=\sum_{y=0}^n \mathbb{P}(X_2=x|X_1=y)\mathbb{P}(X_1=y)$$
If $y<x$, the conditional probability is $0$, thus
$$\mathbb{P}(X_2=x)=\sum_{y=x}^n \mathbb{P}(X_2=x|X_1=y)\mathbb{P}(X_1=y)$$
We can now write
$$\begin{eqnarray}\mathbb{P}(X_2=x|X_1=y) & = & {y \choose x} p^x q^{y-x} \\ \mathbb{P}(X_1=y) & = & {n \choose y} p^y q^{n-y}  \end{eqnarray}$$
and thus
$$\mathbb{P}(X_2=x)=\sum_{y=x}^n {y \choose x} p^x q^{y-x} {n \choose y} p^y q^{n-y}$$
We can work out the combinatorial factors and regroup some other factors
$$\begin{eqnarray}\mathbb{P}(X_2=x) & = & p^x q^{n-x} \sum_{y=x}^n \frac{n!}{x!(y-x)!(n-y)!} p^y \\ & = &\frac{n!}{x!(n-x)!}p^x q^{n-x} \sum_{y=x}^n \frac{(n-x)!}{(y-x)!((n-x)-(y-x))!} p^y \\ & = &\frac{n!}{x!(n-x)!}p^{2x} q^{n-x} \sum_{z=0}^{n-x} \frac{(n-x)!}{(z)!((n-x)-z)!} p^z \end{eqnarray} $$
In the last step, you recognize with the binomium of Newton $(1+p)^{n-x}$, thus
$$\begin{eqnarray}\mathbb{P}(X_2=x) & = & \frac{n!}{x!(n-x)!}p^{2x} q^{n-x} (1+p)^{n-x}\\ & = & \frac{n!}{x!(n-x)!}p^{2x} (1-p)^{n-x} (1+p)^{n-x}\\ & = & \frac{n!}{x!(n-x)!}(p^2)^{x} (1-p^2)^{n-x} \\ \end{eqnarray} $$
This means that $X_2 \sim B(n,p^2)$. I think you can take it from here.

EDIT: For the induction step, assume $X_{k-1}\sim B(n,p^{k-1})$, then
$$\begin{eqnarray}\mathbb{P}(X_{k}=x|X_{k-1}=y) & = & {y \choose x} p^x q^{y-x} \\ \mathbb{P}(X_{k-1}=y) & = & {n \choose y} p^{(k-1)y} (1-p^{k-1})^{n-y}  \end{eqnarray}$$
and thus
$$\mathbb{P}(X_k=x)=\sum_{y=x}^n {y \choose x} p^x q^{y-x} {n \choose y} p^{(k-1)y} (1-p^{k-1})^{n-y}$$
Then work out as before
$$\begin{eqnarray}\mathbb{P}(X_k=x) & = &\frac{n!}{x!(n-x)!}p^x \sum_{y=x}^n \frac{(n-x)!}{(y-x)!((n-x)-(y-x))!} q^{y-x}p^{(k-1)y} (1-p^{k-1})^{n-y} \\& = &\frac{n!}{x!(n-x)!}p^x \sum_{z=0}^{n-x} \frac{(n-x)!}{z!((n-x)-z)!} q^z p^{(k-1)(z+x)} (1-p^{k-1})^{(n-x)-z} \end{eqnarray} $$
In the last step, I again made the change of variables $z=y-x$. And now I can pull out the factor $p^{(k-1)x}$:
$$\begin{eqnarray}\mathbb{P}(X_k=x) & = &\frac{n!}{x!(n-x)!}p^{kx} \sum_{z=0}^{n-x} \frac{(n-x)!}{z!((n-x)-z)!} q^z p^{(k-1)z} (1-p^{k-1})^{(n-x)-z} \\ & = &\frac{n!}{x!(n-x)!}p^{kx} \sum_{z=0}^{n-x} \frac{(n-x)!}{z!((n-x)-z)!} (1-p)^z (p^{k-1})^z (1-p^{k-1})^{(n-x)-z} \\ & = &\frac{n!}{x!(n-x)!}p^{kx} \sum_{z=0}^{n-x} \frac{(n-x)!}{z!((n-x)-z)!} (1-p)^z (p^{k-1})^z (1-p^{k-1})^{(n-x)-z} \\ & = &\frac{n!}{x!(n-x)!}p^{kx} \sum_{z=0}^{n-x} \frac{(n-x)!}{z!((n-x)-z)!} (p^{k-1}-p^k)^z (1-p^{k-1})^{(n-x)-z}\end{eqnarray} $$
Note that the last sum is the binomial expansion of 
$$(1-p^k)^{n-x}=(1-p^{k-1}+p^{k-1}-p^k)^{n-x}$$
We can therefore rewrite
$$\mathbb{P}(X_k=x) = \frac{n!}{x!(n-x)!}(p^k)^x (1-p^k)^{n-x}$$
which confirms that $X_k \sim B(n,p^k)$. This ends the proof by induction.
