Expression for recurrence relation $a_n$ using exponential generating functions $a_0 = 2$, $a_n = na_{n - 1} - n!$ for $n \geq 1$. 
Let $$f(x) = \sum_{n \geq 0}a_n\frac{x^n}{n!}.$$
Multiplying each term in the relation by $\frac{x^n}{n!}$ and summing over values for which the relation is defined and we get: 
$$
\begin{align*}
\sum_{n \geq 1} a_n\frac{x^n}{n!} &= \sum_{n \geq 1}na_{n - 1}\frac{x^n}{n!} - \sum_{n \geq 1}n!\frac{x^n}{n!}\\
f(x) - a_0 &= x\sum_{n \geq 1}a_{n - 1}\frac{x^{n - 1}}{(n - 1)!} - \sum_{n \geq 1}n!\frac{x^n}{n!}\\
f(x) - 2 &= x \cdot f(x) - \left(\frac{1}{1 - x} - 0!\right)\\
f(x) - 2 &= x \cdot f(x) - \frac{x}{1 - x}\\
f(x) - x \cdot f(x) &= 2 - \frac{x}{1 - x}\\
f(x)\left[1 - x\right] &= \frac{2 - 3x}{1 - x}\\
f(x) &= \frac{2 - 3x}{(1 - x)^2}
\end{align*}
$$
After this, we're supposed to find the coefficient of $\frac{x^n}{n!}$ to find a concise expression for $a_n$. However, I've hit several dead ends. Basically I don't know how to play with 
$$f(x) = \frac{2 - 3x}{(1 - x)^2}$$ 
to find the coefficient. Do I use convolution? Partial fraction decomposition? I have no clue. 

I'll take off where you end up. You want:
$\begin{align}
   n! [x^n] \frac{2 - 3 x}{(1 - x)^2}
     &= n! [x^n] (2 - 3 x) (1 - x)^{-2} \\
     &= n! \left(
             2 [x^n] (1- x)^{-2} - 3 [x^{n - 1}] (1 - x)^{-2}
           \right) \\
     &= n! \left(
             2 (-1)^n \binom{-2}{n} - 3 (-1)^{n - 1} \binom{-2}{n - 1}
           \right) \\
     &= n! \left(
             2 \binom{n + 2 - 1}{2 - 1} - 3 \binom{n - 1 + 2 - 1}{2 - 1}
           \right) \\
     &= n! \left(
             2 \binom{n + 1}{1} - 3 \binom{n}{1}
           \right) \\
     &= n! \left(
             2 (n + 1) - 3 n
           \right) \\
     &= - n! (n - 2)
\end{align}$
