The Numerators Diverge Too

It is well known that

$e = \sum_{k=0}^\infty \dfrac{1}{k!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$

What is the value of

$\sum_{k=0}^{\infty} \dfrac{ k+1} {k!} = \dfrac{1}{0!} + \dfrac{2}{ 1!} + \dfrac{3}{2!} + \dfrac{ 4}{3!} + \cdots \quad ?$

$$ Notation: $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$.

Bonus: Prove this by just manipulating the terms from the first identity.