Determining whether ${p^{n-2} \choose k}$ is divisible by $p^{n-k -2}$ for $1 \le k < n$

Let $p$ be an odd prime and $n \ge 3$ a positive integer.

I would like to know whether ${p^{n-2} \choose k}$ is divisible by $p^{n-k -2}$ for $1 \le k < n$. It should be noted that one can show by induction that $n \le p^{n-2}$ when $n \ge 3$ and $p$ is an odd prime (hence ${p^{n-2} \choose k}$ makes sense under these restrictions).

If I have this fact, then I can use the Binomial Theorem to show that the element $(1 + p)$ in the multiplicative group $ \mathbb{Z}/ p^n \mathbb{Z}$ has order at least $p^{n-1}$.

My attempt so far: we have

$${p^{n-2} \choose k} = \frac{p^{n-2}!}{k! (p^{n-2}-k)!}.$$

The goal is to figure out how many copies of $p$ are left over in the numerator (hopefully at least $n-k -2$ copies) after we cancel all the factors of the denominator. However, I am having a hard time identifying the power of $p$ that we will have in the denominator.

Any hints or solutions are greatly appreciated.

Let $d$ be the exponent of the largest power of $p$ which divides a number $n\in\mathbf{N}^{*}$. Let $n = n_0 + n_1 p + \ldots + n_r p^r$ be the expansion of $n$ in base $p$. There are $\left\lfloor \frac{n}{p^k} \right\rfloor $ integers $m\leq n$ such that $p^k | m$ and $\left\lfloor \frac{n}{p^k} \right\rfloor - \left\lfloor \frac{n}{p^{k+1}} \right\rfloor$ integers $m\leq n$ such that $p^k | m$ and $p^{k+1} \nmid m$. Let $e$ be the exponent of the largest power of $p$ which divides $n!$. Then what preceeds shows that $$e = \sum_{k\geq 1} k \left( \left\lfloor \frac{n}{p^k} \right\rfloor - \left\lfloor \frac{n}{p^{k+1}} \right\rfloor \right) = \sum_{k\geq 1} \left\lfloor \frac{n}{p^k} \right\rfloor $$ and as $\left\lfloor \frac{n}{p^k} \right\rfloor = \sum_{i\geq k} n_i p^{i - k}$ we get that $$ e = \frac{n - \sum_{i=0}^r n_i}{p-1} .$$ Noting $v_p (n)$ the exponent of the largest power of $p$ which divides a number $n\in\mathbf{N}^{*}$ we see that $v_p (ab) = v_p(a) v_p(b)$ for $a,b\in \mathbf{N}^{*}$. Note that $v_p ( p^n - j ) = v_p (j)$ when $1\leq j \leq p^n$. Now, $$ k! {p^n \choose k} = p^n (p^n - 1) \ldots (p^n - (k-1))$$ and taking $v_p$ on both sides gives $$ v_p (k!) + v_p\left( {p^n \choose k} \right) = v_p (p^n) + \underbrace{v_p (p^n-1) + \ldots +v_p(p^n - (k-1))}_{= v_p(1) + \ldots + v_p(k-1)=v_p(1\times\ldots\times(k-1)) = v_p((k-1)!)} = n + v_p((k-1)!)$$ that is $$ v_p (k!) + v_p\left( {p^n \choose k} \right) = n + v_p((k-1)!) $$ which shows that $$ v_p\left( {p^n \choose k} \right) = n - v_p (k).$$ Applying this to your case gives $$ v_p\left( {p^{n-2} \choose k} \right) = n - 2 -v_p(k) $$ and to conclude it suffices to show that $v_p(k) \leq k$, because if we show this, then we will have $$ v_p\left( {p^{n-2} \choose k} \right) \geq n - 2 -k $$ which means that $p^{n - 2 -k}$ divides ${p^{n-2} \choose k}$. So we only have to show that $v_p(k) \leq k$. Do you see how to do this ?