Convergence of $\sum \frac{\lambda^y}{y!}$ to $e^\lambda$ How can I show that 
$$ \sum_{y=0}^\infty \frac{λ^y}{y!} = e^λ $$?
I only get so far as to show that the series will converge (by ratio test), but not further, I'm not even sure where to start. 
 A: Your series is one of the possible definitions of the exponential functions in the real field. There are really many possible definitions. I resume here the most known:
1) as a limit:
$$
e^x=\lim_{n \to \infty}\left(1+\dfrac{x}{n} \right)^n  
$$
2) as the solution of differential function: $e^x = f(x)$ such that:
$$ 
f'(x)=f(x)\qquad \mbox{ with the initial condition } \qquad f(0)=1
$$
3) as the solution $e^x=f(x)$ of the functional equation:
$$
f(x+y)=f(x)f(y)\qquad \mbox{ with the condition } \qquad f'(0)=1
$$
4) as a series as in your question:
$$
e^x= 1+\sum_{k=1}^\infty\dfrac{x^k}{k!}
$$
5) as a continued fraction$$
e^x=\cfrac{1}{1-\cfrac{x}{1+x-\cfrac{x}{2+x-\cfrac{2x}{3+x-\cfrac{3x}{4+x-\ddots}}}}}
$$
And we can show that all these definitions are equivalent ( in $\mathbb{R}$).
As an example, using teh binomial formula, we have:
$$
\left(1+\dfrac{x}{n} \right)^n  =1+\sum_{k=1}^n\binom{n}{k}\dfrac{x^k}{n^k}
$$
and the coefficient of the $k-$power is:
$$
\dfrac{n!}{k!(n-k)!\,n^k}=\dfrac {1}{k!} \, \dfrac{1\times 2 \times 3 \cdots \times n}{[1\times 2  \cdots \times (n-k)]\,\times \,\underbrace {n\times n \cdots \times n}_{k \,\mbox{times}}}
$$ 
that, simplifyng and reordering becomes:
$$
\begin{split}
\dfrac{n!}{k!(n-k)!\,n^k}&= \dfrac {1}{k!} \,\dfrac{\overbrace{(n-k+1)\times(n-k+ 2) \cdots \times (n-1)}^{(k-1)\,\mbox{factors}}}{\underbrace {n\times n \cdots \times n}_{(k-1) \,\mbox{factors}}} \\
&=\dfrac {1}{k!}\,\left(1-\dfrac{k-1}{n}\right)\,\left(1-\dfrac{k-2}{n}\right)\cdots\,\left(1-\dfrac{1}{n}\right) 
\end{split}
$$
and taking the limit $n\rightarrow \infty$ we have:
$$
\lim_{n \to \infty}\dfrac{n!}{k!(n-k)!n^k}=\dfrac{1}{k!}
$$
so that:
$$
\exp(x)=\lim_{n \to \infty}\left(1+\dfrac{x}{n} \right)^n = 1+\sum_{k=1}^\infty\dfrac{x^k}{k!}
$$
If you want link the definitions 2) and 4), than derive term by term the series, as suggested in the comments.
