How $\sum_{r=m}^{\infty}\frac{e^{-\lambda}\lambda^r}{r!}=\int_{0}^{\lambda}\frac{e^{-u}u^{m-1}}{(m-1)!}du$

How
$\sum_{r=m}^{\infty}\frac{e^{-\lambda}\lambda^r}{r!}=\int_{0}^{\lambda}\frac{e^{-u}u^{m-1}}{(m-1)!}du$

$$P(X\geq m)=\sum_{r=m}^{\infty}\frac{e^{-\lambda}\lambda^r}{r!};m=0,1,...$$
Show that for any $m=1,2,...$
$$P(X\geq m)=\int_{0}^{\lambda}\frac{e^{-u}u^{m-1}}{(m-1)!}du$$
I couldn't derive it also don't understand why $m$ starts from $1$? why
not from $0$ ?