Why is this true - easy question concerning asymptotics of exponential

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Suppose $\lambda > 0$ is constant as $t \searrow 0$.

In my lecture notes it is written that

$\left(1+\sum_{k=1}^{\infty} \frac{(-\lambda t)^k}{k!}\right) \lambda t = \lambda t + o(t)$

and

$\sum_{k=1}^{\infty} e^{-\lambda t} \frac{(\lambda t)^k}{k!}=o(t)$

as $t \searrow 0$.

Can you explain to me why this is true or tell me some standard rules to deduce such results? I thought it would have to do something with Taylor series, but I am not sure about it. Thank you very much!

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Hint: $$\sum_{k\geq1}\frac{\left(-\lambda t\right)^{k}}{k!}=\sum_{k\geq0}\frac{\left(-\lambda t\right)^{k}}{k!}-1=e^{-\lambda t}-1 $$ and $$\frac{\lambda te^{-\lambda t}-\lambda t}{t}=\lambda e^{-\lambda t}-\lambda\rightarrow0. $$ as $t\rightarrow 0.$