I know the 'Eisenstein's criterion'.
I know that it states "($p\nmid a_{n}$), [$p\mid a_i$ for ($0\le i\le n-1$)], ($p^2\nmid a_0$)".
I know regular way, but I hope to second pf way.
Let $f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}$, deg $f(x)\ge1$.
For some prime $p$, ($p\nmid a_0$) and $p\mid a_i$ for ($1\le i \le n$), ($p^2\nmid a_n$).
Then $f(x)$ is irreducible over $\Bbb Q$.
My Proof. Suppose that $f(x)$ is reducible over $\Bbb Q$.
Then $f(x)=g(x)h(x)$ for some $g(x),h(x) \in \Bbb Z[x]$ (by Gauss lemma), where $g(x)=b_{0}+b_{1}x+\cdots +b_{r}x^{r}$, $\deg g(x)\ge1$, $h(x)=c_{0}+c_{1}x+\cdots +c_{s}x^{s}$, $\deg h(x)\ge1$.
So, $$f(x)=g(x)h(x)=(b_{0}+b_{1}x+\cdots +b_{r}x^{r})(c_{0}+c_{1}x+\cdots +c_{s}x^{s})=b_0c_0+\cdots+b_r c_sx^{r+s}.$$
Then $a_0=b_0 c_0$ and $a_n=b_r c_s$.
Since ($p\nmid a_0$) and $p\mid a_i$ for ($1\le i \le n$), ($p^2\nmid a_n$),
($p\nmid b_0 c_0$) => ($p\nmid b_0$) and ($p\nmid c_0$).
($p^2\nmid b_r c_s$) => either ($p\nmid b_r$) or ($p\nmid c_s$)
Say $p\mid b_r$ and ($p\nmid c_s$).
Let $t$ be the largest integer s.t. ($p\nmid b_t$) and $p\mid b_i$ for ($ 0\lt t \lt i$).
Why largest integer? I thinks set $t$ is smallest integer...
then $a_t=(b_0 c_t +\cdots +b_{t-1} c_1)+b_t c_0$.
Since $p\mid a_t$, then $p\mid b_t c_0$ -Contradiction-
$p\mid b_0 c_t +\cdots +b_{t-1} c_1$ is true? Why?
Set ($p\nmid b_t$) and $p\mid b_i$ for ($ 0\lt t \lt i$).
It means that ($p\mid b_{t+1}$, $p\mid b_{t+2}$, $p\mid b_{t+3}$, $\dots$).
I don't know it... plz help me!
It seems you want to show the Eisenstein criterion with the roles of $a_0$ and $a_n$ swapped? Note that one form follows from the other by noting that $x^{\deg f}f(\frac1x)$ is irreducible iff $f(x)$ is.