Question: I've heard it a lot that Probability is a branch of Statistics(which isn't a branch of mathematics) but as far as I know(High School) it appeals more as the branch of Combinatorics though.

So, can you kindly explain the reasoning behind considering Probability as a branch of Statistics and not as Combinatorics(& Maths in general).

Answer: 
First, determine the df of $X$, because you will need it

$$F\_X(x)=\int\_{0}^x1dt=x$$ for $x\in[0,1]$. Now, the proof by induction: 

* Base case: $F\_1(x)=x^1/1!$, you have shown this.
* Induction step: Assume that $F\_n(x)=x^n/n!$ for $x\in [0,1]$. Then you can find the pdf $f\_n(z)$ of $Z\_n$ by differentiating $F\_n$ $$f\_n(x)=\frac{d}{dx}F\_n(x)=\frac{x^{n-1}}{(n-1)!}$$ for $x\in [0,1]$. Then for the distribution function of $Z\_{n+1}$ we have that \begin{align}F\_{n+1}(x)&=P(Z\_{n+1}\le x)=P(Z\_n+X\_{n+1}\le x)=P(X\_{n+1}\le x-Z\_n)\\[0.2cm]&=\int\_{0}^1P(X\_{n+1}\le x-Z\_n\mid Z\_n=z)f\_n(z)dz\\[0.2cm]&=\int\_0^1F\_X(x-z)\frac{z^{n-1}}{(n-1)!}dz=\int\_0^x(x-z)\frac{z^{n-1}}{(n-1)!}dz\\[0.2cm]&=x\int\_0^x\frac{z^{n-1}}{(n-1)!}dz-\int\_0^x\frac{z^n}{(n-1)!}dz\\[0.2cm]&=x\left[\frac{z^n}{n!}\right]\_0^x-\left[\frac{z^{n+1}}{(n-1)!(n+1)}\right]\_0^x\\[0.2cm]&=\frac{x^{n+1}}{n!}-\frac{x^{n+1}}{(n-1)!(n+1)}=\frac{x^{n+1}(n+1-n)}{n+1!}=\frac{x^{n+1}}{(n+1)!}\end{align} for $x\in[0,1]$. Note the change in the integral limits in the end of the third line. This is because $F\_X(x)=(x-z)$ only if $x-z\in [0,1]$. Otherwise, if $x-z<0$, then $F\_X(x-z)=0$.
The base case: $\mathbb{P}
\{Z\_1\le x\}=x^1/1!$. Assume that the result holds for $n-1$. Then noticing that $\mathbb{P}\{Z\_{n-1}<0\}=0$ and $X$'s are independent, for $x\in [0,1]$ we have

$$
\mathbb{P}\{Z\_n\le x\}=\mathbb{E}[\mathbb{P}\{Z\_{n-1}\le x-X\_n\mid X\_n\}]=\int\_{0}^x \frac{(x-z)^{n-1}}{(n-1)!}dz=\frac{x^n}{n!},
$$

and the result holds by induction.