expectation and variance of poisson distribution

Pmf of Poisson Distribution is as follows:

$$f(X=k;lambda)=frac{lambda^k e^{-lambda}}{k!}$$

Our aim is to derive the the expectation of $E(X)$ and the variance $Var(X)$. Given that the formula of expectation:
$$
E(X)=sum_{k=0}^{infty} k frac{lambda^k e^{-lambda }}{k!}
$$

Notice that when $k=0$, the formula is equal to 0, that is:

$$sum_{k=0}^{infty} k frac{lambda^ke^{-lambda}}{k!}Large|_{k=0}=0$$

Then, the formula become as followed:

$$E(X)=sum_{k=1}^{infty} k frac{lambda^ke^{-lambda}}{k!}$$

$$begin{aligned}E(X)&=sum_{k=0}^{infty} k frac{lambda^ke^{-lambda}}{k!}=sum_{k=0}^{infty} frac{lambda^ke^{-lambda}}{(k-1)!}&=sum_{k=0}^{infty} frac{lambda^{k-1}lambda e^{-lambda}}{(k-1)!}&=lambda e^{-lambda}sum_{k=1}^{infty}frac{lambda^{k-1}}{(k-1)!}end{aligned}$$

Now we need take advantage of Taylor Expansion, recall that:

$$e^x=1+x+frac{x^2}{2!}+frac{x^3}{3!}+cdots+frac{x^{k-1}}{(k-1)!}=sum_{k=1}^{infty}frac{x^{k-1}}{(k-1)!}$$

Compare $E(X)$, we can get:

$$E(X)=lambda e^{-lambda}e^lambda=lambda$$

As known that $Var(X)=E(X^2)-(E(x))^2$, we just get $E(X^2)$. Given that:

$$E(X)=sum_{k=1}^{infty} k frac{lambda^ke^{-lambda}}{k!}=lambda$$

we can use this formula to derive the $E(X^2)$,

$$begin{aligned}E(X)=&sum_{k=1}^{infty} k frac{lambda^ke^{-lambda}}{k!}=lambda\Leftrightarrow&sum_{k=1}^{infty} k frac{lambda^k}{k!}=lambda e^{lambda}\Leftrightarrow&frac{partialsum_{k=1}^{infty} k frac{lambda^k}{k!}}{partial lambda}=frac{partial lambda e^{lambda}}{partial lambda}\Leftrightarrow&sum_{k=1}^{infty}k^2frac{lambda^{k-1}}{k!}=e^lambda+lambda e^lambda\Leftrightarrow&sum_{k=1}^{infty}k^2frac{lambda^{k-1}e^{-lambda}}{k!}=1+lambda \Leftrightarrow&sum_{k=1}^{infty}k^2frac{lambda^{k}e^{-lambda}}{k!}=lambda+lambda^2=E(X^2)end{aligned}$$

then,

$$Var(X)=E(X^2)-(E(X))^2=lambda+lambda^2-(lambda)^2=lambda$$

Thus, we have proved that the Expectation and the Variance of Poisson Distribution are both $lambda$