Suppose that $\pi$ is a derangement of $[n]=\{1,\ldots,n\}$. There are two possibilities.
The simpler one is that there is some $k\in[n-1]$ such that $\pi(n)=k$ and $\pi(k)=n$. (In other words, $\pi$ interchanges $k$ and $n$.) Then the restriction of $\pi$ to the set $A_k=[n-1]\setminus\{k\}$ is a derangement of $A_k$. Conversely, I can start with any derangement $\tau$ of $A_k$ and extend it to a unique derangement $\pi$ of $[n]$ by setting $\pi(i)=\tau(i)$ for $i\in A_k$, $\pi(n)=k$, and $\pi(k)=n$. $|A_k|=n-2$, so there are $f_{n-2}$ derangements of $A_k$ and therefore also $f_{n-2}$ derangements of $[n]$ that interchange $n$ and $k$. Finally, there are $n-1$ possible choices for $k$, so there are $(n-1)f_{n-2}$ derangements of $[n]$ that interchange $n$ with some $k\in[n-1]$.
Now we have to count the derangements of $[n]$ that don’t simply interchange $n$ with some $k\in[n-1]$. Let $\pi$ be such a derangement, and let $k=\pi(n)$. We’re assuming that $n\ne\pi(k)$, so there is an $\ell\in[n-1]$ such that $\ell\ne k$ and $\pi(\ell)=n$. In other words, part of the operation of $\pi$ looks like the following diagram:
$$\ldots\overset{\pi}\longrightarrow\ell\overset{\pi}\longrightarrow n\overset{\pi}\longrightarrow k\overset{\pi}\longrightarrow\ldots\;.\tag{1}$$
We can build a derangement $\tau$ of $[n-1]$ by simply cutting $n$ out of the picture and closing up the gap:
$$\ldots\overset{\tau}\longrightarrow\ell\overset{\tau}\longrightarrow k\overset{\tau}\longrightarrow\ldots\;.\tag{2}$$
That is, $\tau(i)=\pi(i)$ if $i\ne\ell$, and $\tau(\ell)=k$. Now suppose that we start with some derangement $\tau$ of $[n-1]$. How many different derangements of $[n]$ will produce $\tau$ when we ‘cut out $n$’? Pick any $\ell\in[n-1]$, and let $k=\tau(\ell)$. Now insert $n$ between $\ell$ and $k$ by defining a derangement $pi$ of $[n]$ as follows: $\pi(i)=\tau(i)$ if $i\ne\ell$, $\pi(\ell)=n$, and $\pi(n)=k$. This just reverses the operation that took us from diagram $(1)$ to diagram $(2)$. There are $n-1$ possible choices for $\ell$, and each one of them yields a different derangement $\pi$ of $[n]$, so there are $n-1$ different derangements of $[n]$ that yield $\tau$ when we ‘cut out $n$’. And there are $f_{n-1}$ possible derangements $\tau$ of $[n-1]$, so there are altogether $(n-1)f_{n-1}$ derangements of $[n]$ that do not interchange $n$ with some $k\in[n-1]$.
Putting the pieces together, we see that
$$f_n=(n-1)f_{n-1}+(n-1)f_{n-2}=(n-1)(f_{n-1}+f_{n-2})\;.$$
We can now use this recurrence relation (together with the initial values $f_1=0$ and $f_2=1$) to prove by induction on $n$ that
$$f_n=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\;.\tag{3}$$
To get the induction started, you need to check that $(3)$ is true for $n=1$ and $n=2$; I’ll leave that to you. For the induction step, suppose that $n\ge 2$, and that
$$f_m=m!\sum_{k=0}^m\frac{(-1)^k}{k!}$$
for all positive integers $m<n$; that’s your induction hypothesis. Then
$$\begin{align*}
f_n&=(n-1)(f_{n-1}+f_{n-2})\\
&=(n-1)\left((n-1)!\sum_{k=0}^{n-1}\frac{(-1)^k}{k!}+(n-2)!\sum_{k=0}^{n-2}\frac{(-1)^k}{k!}\right)\\
&=(n-1)\left((n-1)!\cdot\frac{(-1)^{n-1}}{(n-1)!}+(n-1)!\sum_{k=0}^{n-2}\frac{(-1)^k}{k!}+(n-2)!\sum_{k=0}^{n-2}\frac{(-1)^k}{k!}\right)\\
&=(n-1)\left((-1)^{n-1}+\sum_{k=0}^{n-2}\big((n-1)!+(n-2)!\big)\frac{(-1)^k}{k!}\right)\;.
\end{align*}$$
Now use the fact that $(n-1)!+(n-2)!=n(n-2)!$ (along with a bit more algebra) to simplify this and show that it really is equal to the righthand side of $(3)$.
