Residue of $z^4e^\frac{1}{z-1}$ at $z=1$

Question

I'm interested in the residue of $z^4e^\frac{1}{z-1}$ at $z=1$ Following the Laurent expansion, I believe I get: $$ z^4\sum_{n=0}^\infty \frac{1}{n!} \left(\frac{1}{z-1}\right)^n = z^4\left(1+\frac{1}{z-1} +\frac{1}{2!(z-1)^2}+\frac{1}{3!(z-1)^3}...\right) $$ I have been told through lectures to find the $z^{-1}$ term, however, in this case, I'm not too sure how to proceed

Answer 1

Hint: $z^4=(1+z-1)^4=1+4(z-1)+6(z-1)^2+4(z-1)^3+(z-1)^4$. Multiplying, we get $1/5!+1/3!+1+2+1=501/120=167/40$ as the residue.