Modular equations $$
x^{13}\equiv4\pmod{101}\\x\equiv5^{5^{5^{5}}}\pmod{47\cdot27}
$$
Equations are separate. How should I approach these? Both has something to do with Euler's theorem, I believe, but all my attempts to solve them were futile. I have an idea on the second one, however I'm not sure there is no mistakes.
Is it true that if $y=5^{5^{5}}\pmod{\phi(47\cdot27)}$, then $5^{y}\equiv5^{5^{5^{5}}}\pmod{47\cdot27}$?
 A: It seems I've found a solution for the second equation.
Let's find $5^{5^{5}}\pmod{\phi(47\cdot27)}$ :
$$
\phi(47\cdot27)=\phi(47)\phi(27)=46\cdot18=2^2\cdot3^3\cdot23=828
$$
Our goal is, therefore, to find the solution of the following system
$$
\cases{x\equiv5^{5^{5}}\pmod{4}\\
       x\equiv5^{5^{5}}\pmod{9}\\
       x\equiv5^{5^{5}}\pmod{23}}
$$ 
Using the same trick with each of the three:
$$
\phi(4)=2\Rightarrow 5^{5^{5}}=5^{3125}=(5^2)^{1562}\cdot5^1\equiv5^1\equiv1\pmod{4}\\
\phi(9)=6\Rightarrow 5^{5^{5}}=5^{3125}=(5^6)^{520}\cdot5^5\equiv5^5\equiv2\pmod{9}\\
\phi(23)=22\Rightarrow 5^{5^{5}}=5^{3125}=(5^{22})^{142}\cdot5^1\equiv5^1\equiv5\pmod{23}
$$
After applying CRT to the obtained system:
$x\equiv5^{5^{5}}\equiv281\pmod{828}$
Nice, know we know that 
$5^{5^{5^{5}}}=(5^{828})^n\cdot5^{281}\equiv5^{281}\pmod{47\cdot27} $
The final strokes:
$$
\cases{5^{281}\equiv(5^{46})^6\cdot5^{5}\equiv5^{5}\equiv23\pmod{47}\\
       5^{281}\equiv(5^{18})^{15}\cdot5^{11}\equiv5^{11}\equiv2\pmod{27}}\\
\\
x\equiv5^{281}\equiv164\pmod{1269}
$$
And that's the answer:
$5^{5^{5^{5}}}\equiv5^{281}\equiv164\pmod{47\cdot27}$
