Proving that $\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b$ 
Prove that for all positive integers $b$ that $$\sum_{a=1}^{b} \frac{a \cdot a! \cdot \binom{b}{a}}{b^a} = b.$$

My idea is induction, but I cannot figure stuff out on the inductive step.

This can be reduced to a telescoping sum.
$$
\begin{align}
\sum_{a=1}^b\frac{a\cdot a!\binom{b}{a}}{b^a}
&=\sum_{a=1}^b\frac{[b-(b-a)]\frac{b!}{(b-a)!}}{b^a}\tag{1}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-\sum_{a=1}^{b-1}\frac{\frac{b!}{(b-a-1)!}}{b^a}\tag{2}\\
&=\sum_{a=1}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}-\sum_{a=2}^b\frac{\frac{b!}{(b-a)!}}{b^{a-1}}\tag{3}\\
&=\frac{\frac{b!}{(b-1)!}}{b^{1-1}}\tag{4}\\[12pt]
&=b\tag{5}
\end{align}
$$
Explanation:
$(1)$: $a=b-(b-a)$ and $a!\binom{b}{a}=\frac{b!}{(b-a)!}$
$(2)$: left sum: $\frac{b}{b^a}=\frac1{b^{a-1}}$
$\phantom{\text{(2):}}$ right sum: $(b-a)\frac{b!}{(b-a)!}=\frac{b!}{(b-a-1)!}$, removing the $a=b$ term
$(3)$: right sum: substitute $a\mapsto a-1$
$(4)$: all terms cancel except for the $a=1$ term in the left sum
$(5)$: evaluate
