How to show that $$10^{124}<100!<10^{180}$$ I believe I'm not supposed to use a calculator. I thought of factorizing the powers of $10$ or using the Stirling's approximation, but I still end up with hard mental math in both.
Would Stirling's approximation actually be sufficiently rigorous to show that this inequality holds true? It's an approximation after all.
On
Observe that
$$ 100! > 1^{9} \times 10^{22} \times 32^{69} $$
which will take care of the first inequality after some massaging, and also that
$$ 100! < 10^{9} \times 31^{22} \times 100^{69} $$
which will take care of the second inequality after similar massaging.
ETA: Simple Art's answer gives better bounds, but this might actually be close to what was intended by the problem poser. (That is, the resultant bounds are pretty close to what is asked for.)
Take the $\log:$ (all base 10 here)
$$\log(100!)=\log(1)+\log(2)+\log(3)+\dots+\log(100)\\<\underbrace{\log(10)+\log(10)+\dots}_{10}+\log(20)+\dots\log(100)$$
where we take $10$ of each, intervals of $10$.
$$=10(\log(10)+\log(20)+\dots+\log(100))=10(10\log(10)+\log(10!))<166$$
Thus, we have
$$100!<10^{166}$$
On the other side, we could notice that
$$\log(1)+\log(2)+\log(3)+\dots+\log(100)>\underbrace{\log(1)+\log(1)+\dots}_{10}+\log(10)+\dots+\log(90)$$
Same manner as last time but in the less than side.
$$\log(1)+\log(1)+\dots+\log(10)+\dots+\log(90)=10(9\log(10)+\log(9!))>145$$
Thus, we end up with
Interestingly, a much better bound than required.