Modular Multiplicative Inverse and RSA?
The multiplicative group $Z_m$ of integers modulo $m$ has exactly $\phi(m)$ elements by definition. Thus for any element $a$ of $Z_m$ the equation
$$
a^{\phi(m)}=1~(mod~m)
$$
holds, where $1$ is the multiplicative identity of $Z_m$ (the residue class $km+1$ of all integers congruent to 1 modulo $m$). 

In RSA, $e$ and $d$ are indeed inverses modulo $\phi(\phi(n))$ in the exponent arithmetic, since the order of the multiplicative group $Z_n=Z_{pq}$ is $\phi(n)=\phi(pq)=(p-1)(q-1).$

So the last line of the image is correct and you get
$$
e=d^{\phi(\phi(n))-1}=d^{-1}~(mod~\phi(n))
$$
in the exponent multiplicative group which is what you want, since
$$
M^{ed}~(mod~ n)=M
$$
if and only if
$$
ed~(mod~\phi(n))=1,
$$
if and only if
$$
e=d^{-1}~(mod~\phi(n))
$$
by uniqueness of inverses in any group.