The zeros on the end of the number are caused by factors of $10$.

$2^{57} \times 3^4 \times 5^{53} = 2^4 \times 3^4 \times 2^{53} \times 5^{53} = 2^4 \times 3^4 \times 10^{53} = 6^4\times10^{53}$

Last non-zero digit comes from $6^4$

$6^1=6$

$6^2=36$

$6^3=\ ...6$

$6^4=\ ...6$

This means the last non-zero digit will be a $6$.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.