Show that for every natural number $n$ there is equality - Stirling numbers of the second kind Show that for every natural number $n$ there is equality: 
$$\left\{ n\atop  3\right\} = \frac{3^{n-1}-2^n+1}{2}$$ 
To prove this equality use equality
$${{n+1}\brace{m+1}} = \sum_{k=0}^{n}{n\choose k}{{k}\brace{m}}$$  and binomial theorem.

Could someone help me with this exercise? I don't know how to go about it ... Thank you!

Just in case you learn about EGFs soon we have from the EGF
$${n\brace k} = n! [z^n] \frac{(\exp(z)-1)^k}{k!}$$
that
$${n\brace 3} = n! [z^n] \frac{(\exp(z)-1)^3}{6}
= \frac{1}{6} n! [z^n] (\exp(3z) - 3 \exp(2z) + 3 \exp(z) - 1)
\\ = \frac{1}{6} (3^n - 3 \times 2^n + 3)
= \frac{1}{2} (3^{n-1} - 2^n + 1)$$
where $n\ge 1.$
