I am told that for a prime $p$, natural number $r$. $\Phi_{p^r}(x)=(x^{p^{r-1}})^{p-1}+(x^{p^{r-1}})^{p-2}+\cdots+(x^{p^{r-1}})+1$ and is irreducible (Here $\Phi$ denotes the cyclotomic polynomial). It is said that it comes from the fact that if $x$ is a primitive $p^r$-th root of unity, then $x^{p^{r-1}}$ is a primitive $p$-th power of unity.
But it is not such clear to me, may I please ask for an explicit explaination for why it is the $p^r$-th cyclotomic polynomial and how to see the fact that it is irreducible?(The fact that for a prime $p$, $x^{p-1}+x^{p-2}+\cdots+x+1$ is the cyclotomic polynomial and is irreducible can be used).
If $\omega$ is a $p^r$-primitive root of unity, then $$ \omega = \large e^{\frac{2\pi i}{p^r}k}$$
for some $k$ relatively prime to $p^r$. Note that $k$ is then relatively prime to $p$.
Then $$\omega^{p^{r-1}} =\large e^{\frac{2\pi i}{p^r}kp^{r-1}} = \large e^{\frac{2\pi i}{p}k}.$$
So $\omega^{p^{r-1}}$ is a $p$-primitive root of unity. Then it satisfies the $p^{th}$ cyclotomic polynomial. So
$$(\omega^{p^{r-1}})^{p-1} + (\omega^{p^{r-1}})^{p-2} + \cdots + (\omega^{p^{r-1}}) + 1 = 0.$$
Therefore $\omega$ is certainly a root of the polynomial you listed.
We know that there are $p^{r-1}(p-1)$ primitive $(p^{r-1})^{th}$ roots of unity, and as we have found a polynomial of that degree that has all of them as its roots. Therefore the polynomial you listed must be the $(p^{r-1})^{th}$ cyclotomic polynomial and is therefore irreducible by the general theory of cyclotomic polynomials.