Here $ n,p $ are positive integers with $ p $ being a prime number.
I proved the result below, but I would like to know if there is an alternative way of doing it using reduction modulo $ p $.
Note that $$ f(X+1)=X^{p^{n}}+\sum_{k=1}^{p^{n}-1}\binom{p^{n}}{k}X^{k}+1+p-1=X^{p^{n}}+\sum_{k=1}^{p^{n}-1}\binom{p^{n}}{k}X^{k}+p $$ hence if suffices to prove that $$ \binom{p^{n}}{k} \equiv 0 \text{ modulo } p $$ for any $ k \in \overline{1,p^{n}-1} $ in order to conclude that $ f(X+1) $ is irreducible in $ \Bbb{Q}[X] $ via Eisenstein's criterion.
Recall that for a positive integer $ n $ we define $ v_{p}(n) $ (the p-adic valuation of $ n $ )to be the exponent of the largest power of the prime $ p $ that divides $ n $. We will use Legendre's formula $$ v_{p}(n!)=\sum_{i=1}^{\infty}\left \lfloor \frac{n}{p^{i}} \right \rfloor $$ to show that $ \binom{p^{n}}{k} \equiv 0 \text{ modulo }p $ for any $ k \in \overline{1, p^{n}-1} $.
For any such $ k $ we have that $ k<p^{n} $ so \begin{align*} v_{p}(k!)&=\sum_{i=1}^{\infty} \left \lfloor \frac{k}{p^{i}} \right \rfloor \nonumber \\ &=\sum_{i=1}^{n-1} \left \lfloor \frac{k}{p^{i}} \right \rfloor \nonumber \\ &\le \frac{k}{p}+ \frac{k}{p^{2}}+ \cdots \frac{k}{p^{n-1}} \nonumber \end{align*} Furthermore we have \begin{align*} v_{p}((p^{n}-k)!)&=\sum_{i=1}^{n-1} \left \lfloor \frac{p^{n}-k}{p^{i}} \right \rfloor \nonumber \\ &\le \frac{p^{n}-k}{p}+\frac{p^{n}-k}{p^{2}}+ \cdots + \frac{p^{n}-k}{p^{n-1}} \nonumber \\ &=p+p^{2}+ \cdots +p^{n-1}-\big( \frac{k}{p}+\frac{k}{p^{2}}+ \cdots + \frac{k}{p^{n-1}} \big) \end{align*} Therefore $$ v_{p}(k!)+v_{p}((p^{n}-k)!) \le p+p^{2}+ \cdots + p^{n-1}=\frac{p^{n}-1}{p-1}-1 $$ On the other hand we have that \begin{align*} v_{p}((p^{n})!)&=\sum_{i=1}^{\infty} \left \lfloor \frac{p^{n}}{p^{i}} \right \rfloor \nonumber \\ &=\sum_{i=1}^{n} \left \lfloor \frac{p^{n}}{p^{i}} \right \rfloor \nonumber \\ &=p^{n-1}+p^{n-2}+ \cdots + p+1 \nonumber \\ &=\frac{p^{n}-1}{p-1} \end{align*} So we have proved that $$ v_{p}((p^{n})!)>v_{p}(k!)+v_{p}((p^{n}-k)!) $$ and as $$ \binom{p^{n}}{k}=\frac{(p^{n})!}{k!(p^{n}-k)!} $$ this goes to show that $$ v_{p}\Big(\binom{p^{n}}{k}\Big)=v_{p}((p^{n})!)-\big(v_{p}(k!)+v_{p}((p^{n}-k)!)\big) \geq 1 $$ hence $ p $ divides $ \binom{p^{n}}{k} $.
In conclusion, as aforementioned we have that $ f(X+1) $ is Eisenstein with respect to the prime $ p $ so it is irreducible in $ \Bbb{Q}[X] $ hence $ f(X) $ is irreducible in $ \Bbb{Q}[X] $ and since its content is $ 1 $, it is a primitive polynomial therefore irreducible in $ \Bbb{Z}[X] $ as well.
Now, if we consider $ f(X+1) $ over $ \Bbb{F}_{p} $ we have $$ f(X+1)=(X+1)^{p^{n}}+p-1=X^{p^{n}}+1+p-1=X^{p^{n}} $$ via Frobenius and the later polynomial is clearly reducible over $ \Bbb{F}_{p} $ so I cannot use the reduction modulo $ p $ under this form.
Can someone point out if it is possible though to use a reduction modulo $ p $ argument to prove the irreducibility of $ f $? Thank you!
