Proof of Dobinski's formula for Bell numbers

The Bell number $B(n)$ is defined as $\sum_{k=1}^n S(n,k)$ where $S(n,k)={n\brace k}$ is a Stirling number of the second kind. I would like to learn how to prove the following identity (Dobinski's formula): $$B(n)=\frac{1}{e}\sum_{j=0}^{\infty}\frac{j^n}{j!}.$$

The definition $$ B_n = \sum_{k=0}^{n}{n\brace k} \tag{1}$$ imply that Bell number fulfill the following recurrence relation: $$ B_{n+1} = \sum_{k=0}^{n}\binom{n}{k}B_k \tag{2}$$ leading to their exponential generating function $$ f(x) = \sum_{n\geq 0}\frac{B_n}{n!}x^n = \exp\left(e^x-1\right).\tag{3} $$ Now, since $$ \exp\left(e^x\right) = 1+e^x+\frac{e^{2x}}{2!}+\frac{e^{3x}}{3!}+\ldots \tag{4}$$ the coefficient of $x^n$ in the RHS of $(4)$ is given by $$ \frac{1}{n!}+\frac{2^n}{n!2!}+\frac{3^n}{n!3!}+\ldots\tag{5}$$ and the claim (Dobinski's formula) $$ B_n = \frac{1}{e}\sum_{j\geq 0}\frac{j^n}{j!}\tag{6}$$ readily follows.