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{\partial\sum_{k=1}^{\infty} k \frac{\lambda^k}{k!}}{\partial \lambda}=\frac{\partial \lambda e^{\lambda}}{\partial \lambda}\\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\lambda^{k-1}}{k!}=e^\lambda+\lambda e^\lambda\\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\lambda^{k-1}e^{-\lambda}}{k!}=1+\lambda \\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\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 $\lambda$