Problem
What is the last digit of $235!^{69}$?
It's been far too long since I did any modulo calcuations, and even then, the factorial would set me back.
My initial thought goes to the last digit $5$ in $235$. If $235!$ can be shown to have a $5$ as last digit, raising it to any natural number shouldn't change that, right?
On
If $n>4$, then $(n!)^k \equiv 0\mod{10}$ $\implies$ the last digit is a $0$.
To see this, note since $n>4$, $n!=(n)(n-1)\cdot\cdot\cdot(5)(4)(3)(2)(1)$. We can rewrite this as: $$10\cdot[(n)(n-1)\cdot\cdot\cdot(4)(3)(1)]$$ Now our modulo definition states that since 10 is a factor of $n!$, taking its mod will yield a zero remainder ($\exists h\in \{1,2,3...\}$ s.t. $h=\frac{n!}{10}=\frac{10\cdot[(n)(n-1)\cdot\cdot\cdot(4)(3)(1)]}{10}=[(n)(n-1)\cdot\cdot\cdot(4)(3)(1)]$ which is natural number since the naturals are closed under addition and multiplication). Thus, we have proven $n!$'s congruence to $0\mod10$. Now we simply note the following identity:
If $a\equiv b\mod{m}$, then $a^q\equiv b^q\mod{m}$.
It is zero. Actually $n!$ is a multiple of $10$ for $n\ge 5$, and raising a multiple of $10$ to any natural power results in a multiple of $10$ again.
To elaborate, $235!$ ends in $\lfloor\frac{235}5\rfloor+\lfloor\frac{235}{25}\rfloor+\lfloor\frac{235}{125}\rfloor+\ldots = 57$ zeroes, hence $235!^{69}$ ends in $57\cdot 69=3933$ zeroes. The last non-zero digit of $235!$ turns out to be a $6$ (however, showing that is not that trivial), and for that your argument applies that this does not change by raising to natural powers (because already $6\cdot 6\equiv 6\pmod{10}$. We conclude that the last non-zero digit of $235!^{69}$ is a $6$.