When 0$\leq a\leq$p - 1 show that $ord_p((ap^n)!)$= a$(1+p+p^2+......+p^{n-1})$.
What i know is let n≥1 and n=$n_0+...n_ℓp^ℓ$ be the p-adic expansion of n. Define $\alpha_p(n)=n_0+...+n_ℓ$. Then $ord_p(n!)=\frac{n−\alpha_p(n)}{p−1}$.
And also in the previous exercise $ord_p((p^n)!)$=$\frac{p^n - \alpha_p(p^n)}{p-1}$=$\frac{p^n - 1}{p-1}$=$1+p+p^2+...+p^{n-1}$ (since $p^n$=1×$p^n$ and so $\alpha_p(p^n)$=1)
But I have no idea how to do this exercise. Thank you for your help.
Hint: The $p$-adic expansion of $a p^n$ is ... $a p^n$.
Plug that into the two formulae you claim you know in your second paragraph and you're done. (I hope you know that $\dfrac{p^n -1}{p-1} = 1 + p + ... + p^{n-1}$.)