Expected value of this continuous RV I'm skimming through a basic introductory level stats-book and there's a problem which begins with:

Let $X_n$ be a continuous random variable with a $PDF = f_n(x) = \frac {x^n}{n!} e^{-x}, x>0$.

The full problem is fairly long and in order for it to be solved, the reader needs to find the expected value of $X_n$.
However, the textbook just casually goes on to denote that the expected value of $X_n$ is $n+1$ without providing any information on how that value is actually calculated. 
My question is: how can I directly calculate the expected value, without relying on computer software? 
 A: $$E[X] = \int_0^\infty x\cdot\frac{x^n}{n!}e^{-x}\ dx = \int_0^\infty \frac{x^{n+1}}{n!}e^{-x}\ dx$$
From integration by parts, we have:
$$E[X] = -\frac{x^{n+1}}{n!}e^{-x}\bigg|_0^\infty + (n+1)\int_0^\infty\frac{x^n}{n!}e^{-x}\ dx$$
The first term evaluates to $0$, and the second term is $(n+1)$ times the PDF and we know this integrates to $1$. This leaves us with:
$$E[X] = n+1$$

If you still are looking further for clarification on how the PFD integrates to $1$, you can repeat the above logical process. 
$$\int_0^\infty\frac{x^n}{n!}e^{-x}\ dx = \int_0^\infty \frac{x^{n-1}}{(n-1)!}e^{-x}\ dx = \dots = \int_0^\infty xe^{-x}\ dx = \int_0^\infty e^{-x}\ dx = 1$$
A: If you accept as a given fact that for each $n > 0$ that the function $$f_n(x) = \frac{x^n}{n!} e^{-x}, \quad x > 0$$ is a probability density function satisfying $$\int_{x=0}^\infty f_n(x) \, dx = 1,$$ then the expectation is trivial:  $$\begin{align*} \operatorname{E}[X] &= \int_{x=0}^\infty x f_n(x) \, dx \\ &= \int_{x=0}^\infty \frac{x^{n+1}}{n!} e^{-x} \, dx \\ &= \frac{(n+1)!}{n!} \int_{x=0}^\infty \frac{x^{n+1}}{(n+1)!} e^{-x} \, dx \\ &= (n+1) \int_{x=0}^\infty f_{n+1}(x) \, dx \\ &= n+1. \end{align*}$$
