Both sides of the identity are just the expected number of subsets $A \subseteq [n]$ such that $\max_{i \in A}\sigma(i) < \sigma(n+1)$ in a random permutation $\sigma$ of $[n+1]$ counted in two different ways.
To be more precise (and to avoid mentioning the notions of probability theory, which confuse first year students sometimes), let us denote by $N$ the number of pairs $(\sigma, A)$, where $A$ is a subset of $[n]:=\{1,\dots,n\}$, $\sigma$ is an element of $S_{n+1}$, i.e. a permutation of $[n+1]$, and $\max_{i \in A}\sigma(i) < \sigma(n+1)$.
One the one hand, observe that
\begin{align}
N = & \ \sum_{A \subseteq [n]} \#\left\{ \sigma \in S_{n+1} : \max_{i \in A}\sigma(i) < \sigma(n+1) \right\} \\
= \sum_{k=0}^{n}&\sum_{\substack{A \subseteq [n] \\ |A|=k}} \#\left\{ \sigma \in S_{n+1} : \max_{i \in A}\sigma(i) < \sigma(n+1) \right\} = \sum_{k=0}^{n} \binom{n}{k}\frac{(n+1)!}{k+1}
\end{align}
since there are $\binom{n}{k}$ ways to fix a $k$-element $A\subseteq [n]$ and for each of them, there are $\frac{(n+1)!}{k+1}$ ways to choose $\sigma \in S_{n+1}$ with the desired property.
On the other hand, note that
\begin{align}
N = \sum_{\sigma \in S_{n+1}}\ &\#\left\{ A \subseteq [n] : \max_{i \in A}\sigma(i) < \sigma(n+1) \right\} \\
= \sum_{m=1}^{n+1}\sum_{\substack{\sigma \in S_{n+1} \\ \sigma(n+1)=m}}\!\! &\#\left\{ A \subseteq [n] : \max_{i \in A}\sigma(i) < \sigma(n+1) \right\} = \sum_{m=1}^{n+1} n! \cdot 2^{m-1} = n! \cdot (2^{n+1}-1)
\end{align}
since given $m \in [n+1]$, there are $n!$ ways to fix $\sigma \in S_{n+1}$ such that $\sigma(n+1)=m$ and for each of them, there are $2^{m-1}$ ways to choose $A \subseteq [n]$ with the desired property.
Dividing both the expressions for $N$ found above by $(n+1)!$, we obtain the desired identity.
Remark 1. One can prove more general identity in the same vain. Given $x \in \mathbb{R}$, consider a random variable $f$ defined by the equation
$$f(\sigma)= \sum_{\substack{A \subseteq [n] \\ \max_{i \in A}\sigma(i) < \sigma(n+1)}} x^{|A|+1}$$
for a permutation $\sigma \in S_{n+1}$ chosen uniformly at random. Similar to how we handled the $x=1$ case above, one can count the expectation of $f$ in two straightforward ways to conclude that
$$ \sum_{k=0}^{n} \binom{n}{k}\frac{x^{k+1}}{k+1} = \mbox{E}[f] = \sum_{m=1}^{n+1} \frac{x(1+x)^{m-1}}{n+1} = \frac{(1+x)^{n+1}-1}{n+1}.$$
Of course there are many simpler non-combinatorial ways to prove this identity.
Remark 2. In is known than any $k+1$ elements of $[n+1]$ lie on the same cycle in a random permutation of $[n+1]$ with probability $\frac{1}{k+1}$. Therefore, if one finds it more comprehensible, there would be no harm to any of the arguments above in replacing all the '$\max_{i \in A}\sigma(i) < \sigma(n+1)$' conditions with '$A$ and $n+1$ lie on the same cycle in $\sigma$'.
