Why does decryption in RSA not fail with 2 different moduli?
The question assumes, wrongly, that $a^b\bmod n$ is the same as $a^{b\bmod n}$ or $a^{b\bmod n}\bmod n$. In general these are three different quantities, as demonstrated with $a=5$, $b=5$, $n=3$:
$$\begin{array}{}
a^b\bmod n&amp;=&amp;\left(5^5\right)\bmod 3&amp;=&amp;3125\bmod 3&amp;=&amp;2\\
a^{b\bmod n}&amp;=&amp;5^{(5\bmod 3)}&amp;=&amp;5^2&amp;=&amp;25\\
a^{b\bmod n}\bmod n&amp;=&amp;\left(5^{(5\bmod 3)}\right)\bmod 3&amp;=&amp;25\bmod3&amp;=&amp;1
\end{array}
$$
Note: going from the second to third expression of the last line uses the result obtained above.