mgf of a poisson distribution -- what am I doing wrong? I'm not sure what I'm doing wrong here, can someone please explain? Thank you so much in advance!

 A: Everything is correct up to the step "Plug in":  $$M(t) = \sum_{k=0}^\infty e^{tk} \frac{e^{-\lambda} \lambda^k}{k!}.$$  Then you drop the summation symbol, and from there, things deteriorate rapidly.  You can't just arbitrarily plug in a value for $k$, when $k$ is an index of summation over the nonnegative integers.  Rather, you should write $$M(t) = \sum_{k=0}^\infty e^{-\lambda} \frac{(e^t \lambda)^k}{k!}.$$  Now, let $\lambda^* = e^t \lambda$, so that you get $$M(t) = \sum_{k=0}^\infty e^{-\lambda} \frac{(\lambda^*)^k}{k!}.$$  If the term $e^{-\lambda}$ were instead $e^{-\lambda^*}$, then this sum would be $1$, because it is a sum of the probability mass function of a Poisson random variable with parameter $\lambda^*$ over its support.  So, how could we adjust this expression?  It's simple:
$$\begin{align*} M(t) &= \sum_{k=0}^\infty e^{-\lambda} \frac{(\lambda^*)^k}{k!} \\
&= \sum_{k=0}^\infty e^{-\lambda} e^{\lambda^*} e^{-\lambda^*} \frac{(\lambda^*)^k}{k!} \\
&= e^{-\lambda} e^{\lambda^*} \sum_{k=0}^\infty e^{-\lambda^*}\frac{(\lambda^*)^k}{k!} \\
&= e^{\lambda^*} e^{-\lambda} \\
&= e^{\lambda^* - \lambda} \\
&= e^{e^t \lambda - \lambda} \\
&= e^{(e^t - 1)\lambda}. \end{align*}$$  This works because $\lambda^*$ is not a function of the index of summation $k$, so we are free to multiply and divide by this factor as needed, much like how we might write $$\frac{x}{\sqrt{x-1}} = \frac{x \sqrt{x-1}}{\sqrt{x-1} \sqrt{x-1}} = \frac{x\sqrt{x-1}}{|x-1|}.$$  And as you can see, $e^{-\lambda^*} e^{\lambda^*} = 1$, so we have not changed the value of the sum when we inserted this factor into the sum in the second step.  Then, moving to the third step, we pull out the constant (with respect to $k$) factor $e^{-\lambda}e^{\lambda^*}$ from the sum.
A: One "typo" is where you write "for $x=1,2,3,\ldots$", but you need "for $x=0,1,2,3,\ldots$", starting with $0$ rather than with $1$.
Just why you plugged in $k=0$ and expected to get the answer from that is a mystery to me.
You correctly wrote:
$$
M(t) = \sum_{k=0}^\infty e^{tk} e^{-\lambda} \frac{\lambda^k}{k!}. \tag 1
$$
Notice that in $(1)$, the factor $e^{-\lambda}$ does not change as $k$ goes from $0$ to $\infty$ and therefore can be pulled out:
$$
M(t) = e^{-\lambda} \sum_{k=0}^\infty e^{tk} \frac{\lambda^k}{k!}.
$$
The last sum can be written as
$$
\sum_{k=0}^\infty\frac{(e^t \lambda)^k}{k!}. \tag 2
$$
Now remember a basic result from calculus:
$$
e^a = \sum_{k=0}^\infty \frac{a^k}{k!}. \tag 3
$$
Just apply $(3)$ with $a= e^t \lambda$, and that tells you the value of the sum in $(2)$, and then consequently that of the sum in $(1)$.
