Problem: Find all nonnegative integer solutions to \[ 2^a + 3^b + 5^c = n!. \]

Solution: For $n \le 4$, one can check the only solutions are:
\begin{align*}
  2^2 + 3^0 + 5^0 &= 3! \\
  2^1 + 3^1 + 5^0 &= 3! \\
  2^4 + 3^1 + 5^1 &= 4!.
\end{align*}
Now we prove there are no solutions for $n \ge 5$.

A tricky way to do this is to take modulo $120$, since
\begin{align*}
  2^a \pmod{120} &\in \{ 1, 2, 4, 8, 16, 32, 64 \} \\
  3^b \pmod{120} &\in \{ 1, 3, 9, 27, 81 \} \\
  5^c \pmod{120} &\in \{ 1, 5, 25 \}
\end{align*}
and by inspection one notes that no three
elements have vanishing sum modulo $120$.

I expect most solutions to instead use casework.
Here is one possible approach with cases (with $n \ge 5$).
First, we analyze the cases where $a < 3$:
\begin{itemize}
  \ii $a=0$: No solutions for parity reasons.
  \ii $a=1$: since $3^b + 5^c \equiv 6 \pmod 8$,
  we find $b$ even and $c$ odd (hence $c \neq 0$).
  Now looking modulo $5$ gives that $3^b + 5^c \equiv 3 \pmod 5$,
  \ii $a=2$: From $3^b + 5^c \equiv 4 \pmod 8$,
  we find $b$ is odd and $c$ is even.
  Now looking modulo $5$ gives a contradiction,
  even if $c = 0$,
  since $3^b \in \{2,3 \pmod 5\}$ but $3^b + 5^c \equiv 1 \pmod 5$.
\end{itemize}
Henceforth assume $a \ge 3$.
Next, by taking modulo $8$ we have $3^b+5^c \equiv 0 \pmod 8$,
which forces both $b$ and $c$ to be odd (in particular, $b,c > 0$).
We now have
\begin{align*}
  2^a + 5^c &\equiv 0 \pmod 3 \\
  2^a + 3^b &\equiv 0 \pmod 5.
\end{align*}
The first equation implies $a$ is even,
but the second equation requires $a$ to be odd, contradiction.
Hence no solutions with $n \ge 5$.