Pick $\sigma_1$. That determines that there will be $\sigma_1-1\in\{0,\dotsc,n-1\}$ inversions involving $\sigma_1$, no matter what other values you will pick. The options you have for inversions involving only the remaining values don't depend on the value of $\sigma_1$ you picked; you can imagine the remaining values "compressed" into the range $\{2,\dotsc,n\}$ by moving the values below $\sigma_1$ up by one to fill the gap; that doesn't change the ordering of the values. So the options remaining are the ones for a permutation of $n-1$ objects. Thus the formula follows by induction, since the options $0,\dotsc,n-1$ for $\sigma_1-1$ add a factor $1+x+\dotso+x^{n-1}$ to the generating function for $S_{n-1}$.
Just to get the question clear: The descent set of a permutation $\pi$ is defined as the points where a descent occurs, namely $\operatorname{Des}(\pi) = \{i: \pi_i > \pi_{i+1}\}$.
The first descent in a permutation of size $n$ is defined as $F(\pi) = \min (\operatorname{Des}(\pi) \cup \{n\})$. (The union with $\{n\}$ is just to cover the case of the permutation with no descents, namely $(1, 2, \dots, n)$, for which define the first descent to be at the last position $n$.)
A "desarrangement" is defined as a permutation $\pi$ for which $F(\pi)$ is even.
We want to find the exponential generating function for $E_n$, the number of desarrangements of length $n$.
Method 1.
Prove that the number of desarrangements is the same as the number of derangements, via a bijection. As this is the approach Brian M. Scott alludes to in his answer, I will not elaborate on this further. (Besides, it requires you to know / derive the EGF of the number of derangements.)
Method 2.
We'll count the number of permutations $\pi$ of size $n$ that have $F(\pi) = 2k$, for each $k$.
To have $F(\pi) = 2k$, if $2k < n$, it means that the relative order of the first $2k+1$ elements is such that they are all ascents, except for a final descent at position $2k$. The probability (over uniform permutations of size $n$) that this happens is $\dfrac{2k}{(2k+1)!}$, as of the $(2k+1)!$ possible orderings of the first $(2k+1)$ elements, only $2k$ have this property. (Essentially, arrange all $(2k+1)$ of them in increasing order, and then move any of the first $2k$ non-maximum elements to the end.)
The other way in which we can have $F(\pi) = 2k$, per our definition is if $n = 2k$ and the permutation is $(1, 2, \dots, n)$.
So the number of permutations of size $n$ with $F(\pi)$ even is
$$E_n = \sum_{0 \le 2k < n} n!\frac{2k}{(2k+1)!} + [n\text{ is even}]$$
The EGF is
$$
\begin{align}
E(z) = \sum_{n\ge 0} E_n \frac{z^n}{n!}
&= \sum_{n\ge0}\sum_{0\le 2k < n} n!\frac{2k}{(2k+1)!} \frac{z^n}{n!} + \sum_{n \ge 0} [n\text{ is even}]\frac{z^n}{n!} \\
&= \sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}\sum_{n>2k} z^n \right) + \frac{e^z + e^{-z}}2 \\
&= \sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}\frac{z^{2k+1}}{1-z} \right) + \frac{e^z + e^{-z}}2 \\
&= \frac{1}{1-z}\sum_{k\ge 0}\left(\frac{2k}{(2k+1)!}z^{2k+1}\right) + \frac{e^z + e^{-z}}2 \\
\end{align}
$$
Now to evaluate the sum above, we start with the known
$$\frac{e^z - e^{-z}}{2} = \sum_{k\ge 0} \frac{z^{2k+1}}{(2k+1)!}$$
and differentiate it:
$$\frac{e^z + e^{-z}}{2} = \sum_{k \ge 0} \frac{(2k+1)z^{2k}}{(2k+1)!}$$
so
$$\begin{align}
\sum_{k \ge 0} \frac{2k}{(2k+1)!} z^{2k+1}
&= \frac{z(e^z + e^{-z})}{2} - \sum_{k \ge 0} \frac{z^{2k+1}}{(2k+1)!} \\
&= \frac{z(e^z + e^{-z})}{2} - \frac{e^{z} - e^{-z}}{2} \\
&= \frac{e^{z}(z-1) + e^{-z}(z+1)}{2}
\end{align}$$
which makes our generating function
$$
\begin{align}
E(z)
&= \frac{1}{1-z}\frac{e^{z}(z-1) + e^{-z}(z+1)}{2} + \frac{e^z + e^{-z}}2 \\
&= \frac{e^{-z}}{2}\left(\frac{z+1}{1-z} + 1\right) \\
&= \frac{e^{-z}}{1-z}
\end{align}
$$
(which unsurprisingly happens to be the same as $D(z)$ the exponential generating function for derangements).
Method 3.
We can use the "symbolic method" from the book Analytic Combinatorics (available online!) by Flajolet and Sedgewick.
Let $\mathcal{H}$ denote the class of "hooks" (or "hockey sticks"), by which I mean permutations that are increasing except for the last element. So it's a set of numbers, arranged in increasing order, followed by some number that is not the maximum. In the notation of the book,
$$\mathcal{H} = \operatorname{S\scriptsize ET}\left(\mathcal{Z}\right) ^\blacksquare\star \mathcal{Z}$$
immediately giving that the EGF of $\mathcal{H}$ is
$$H(z) = \int_0^z (\frac{d}{dt}e^t) t \, dt = e^z(z-1) + 1$$
(You could of course get the same with actual enumeration, using $H_n = (n-1)$ (you only have choice of the last element), but we're trying to avoid explicit counting in this method.)
Let $\mathcal{E}$ be the class of desarrangements, that we want to count. A desarrangement is either a member of $\mathcal{H}$ of odd size (which therefore has its descent in an even position), followed by an arbitrary permutation (and then relabelled), or an increasing permutation (a set of numbers arranged in increasing order) of even size. In notation,
$$\mathcal{E} = \mathcal{H}_{\text{odd}} \star \operatorname{S\scriptsize EQ}(\mathcal{Z}) + \operatorname{S\scriptsize ET}_{\text{even}}(\mathcal{Z})$$
which immediately translates to the EGF
$$\begin{align}
E(z)
&= \frac{H(z) - H(-z)}{2} \frac{1}{1-z} + \frac{e^z + e^{-z}}{2} \\
&= \frac{e^z(z-1) + 1 - e^{-z}(-z-1) - 1}{2}\frac{1}{1-z} + \frac{e^z + e^{-z}}{2}\\
&= \frac{e^{-z}}{1-z}
\end{align}
$$
as before. Even some intermediate expressions turned out the same as in the previous approach, but the difference this time is that we argued "globally" (which correponds to multiplying generating functions etc.) instead of having to count $E_n$ separately for every $n$.
Best Answer
You are looking at permutations of $n$ and each cycle contributes a factor $x$.
Now, when you place the first element, there is only one possibility, it starts a cycle, this gives $x$. When you place the second element, it either starts a new cycle giving $x$ or it is placed in the cycle of the first element, this gives $x+1$ When you place element $k$, it either starts a new cycle giving $x$ or it is placed as image of one of the elements that are already there, this gives $x+k-1$.
So, in total, you get the product on the right-hand side.