Why are these two equivalent? (Modular multiplicative inverse) According to Wikipedia's entry "Modular Multiplicative Inverse," $d\equiv e^{-1}  \pmod {\phi(n)}$ and $ed\equiv 1 \pmod{\phi(n)}$ are equivalent. Why is this the case? Can someone provide a step-by-step explanation as to how you can go from the first expression to the second? (I'm just starting to learn modulos to understand RSA, please keep it at a beginner level).
 A: $$
\begin{align}
d & \equiv e^{-1}  \pmod {\phi(n)} \\ \\
ed & \equiv ee^{-1} \pmod{\phi(n)} \\ \\
ed & \equiv 1 \pmod{\phi(n)}
\end{align}
$$
$e^{-1} \pmod{\phi(n)}$ denotes (and presumes the existence of) the multiplicative inverse of the element, $e \pmod {\phi(n)},\;$; and (assuming it exists), the multiplicative inverse of $\,e\,$ is denoted $e^{-1}$. 
By definition, $e^{-1}$ exists and is the multiplicative inverse of $\,e\,$ if and only if satisfies $$ee^{-1} \equiv e^{-1}e \equiv 1 \pmod{\phi(n)}$$ where $1$ denotes the multiplicative identity.
A: That's what "$e^{-1}$" means.
The number (or equivalence class of numbers) called $e^{-1}$ is by definition that which when multiplied by $e$ gives $1$.
For example, $2\cdot6\equiv1\mod{11}$; therefore $2=6^{-1}$ in the field of integers modulo $11$.
Later note: Since the issue was raised in comments: For some values of $n$ and $e$, the number $e$ has no mod-$n$ multiplicative inverse.  That doesn't alter the validity of what I wrote above.  In those cases in which there exists a number $d$ such that $ed\equiv1\pmod n$, the number $d$ is called $e^{-1}$.
Of course, for all this to make sense, it must have been shown that multiplication actually exists in this context, i.e. if $a_1\equiv a_2$ and $b_1\equiv b_2$, then $a_1b_1\equiv a_2b_2$.
A: The number $e^{-1}$ (defined specifically modulo $\varphi(n)$) is an integer $f$ such that $ef \equiv 1 \mod \varphi(n)$, i.e. $ef = 1 + k\varphi(n)$ for some $k$. (It might help you to computer $5^{-1} \mod 7$ or something. You should get many answers, and they should all be integers. There should only be one answer between $0$ and $6$ - what is it?)
Your two statements, written in "non-modular" arithmetic, are:
(a) $d = f + \ell\varphi(n)$ for some $\ell$
(b) $ed = 1 + m\varphi(n)$ for some $m$.
To prove that (a) implies (b), try multiplying both sides by $e$. To prove that (b) implies (a), try multiplying both sides by $f$.
