I Need Scissors 61!

$5!$ will already give us $120$ . Any positive integers we multiply by $120$ will have a unit digit of $0$

Therefore $61!$ will have a unit digit of $\color{#BA33D6}{\boxed{0}}$

Very easy.

the last digit of every number $\text{ n > 4 }$ always is $\boxed{0}$

$\therefore{}$ the last digit of $\text{61!}$ is $\boxed{0}$ .

$61!= 1\times 2 \times 3 \times 4 \times 5 \times \cdots \times 61$

$2 \vert 61!, 5 \vert 61! \Rightarrow 10 \vert 61!$

The last digit is therfore $\boxed{0}$