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Consider $f(x) = p^{n-1}x^n - 1 \in \mathbb{Q}[x]$. I want to show that it's irreducible when $p$ is prime.

Neither reduction of the coefficients modulo some prime nor Eisenstein seems to work here.

I also have had no luck finding a suitable substitution (such as $f(x+1)$) to apply one of these tests to. I may be miscalculating something, but I'd appreciate some help.

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If you substitute $x$ for $x/p$, and then you multiply by $p$ in order not to have denominators, then you get the polynomial $x^n-p$. But then Eisenstein's criterion for the prime $p$ gives you the irreducibility.

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Warning: This method is quite overkill, see xarles' answer for simple method.

You can consider reverted polynomial $x^nf(1/x)=-x^n+p^{n-1}$. This polynomial will be irreducible iff the original polynomial is irreducible (see Prove that the polynomial $x^nf(1/x)$ with reverted coefficients is also irreducible polynomial over $\mathbb{Q}$ ). To show this final polynomial is irreducible, Newton polygons can be used (see Explaining Newton Polygon for proving irreducibility of polynomial in $\mathbb{Z}[x]$ in elementary way).

Basically if you construct Newton polygon corresponding to $p$, you will get line segment between $[0,n-1]$ and $[n,0]$. Since this line does not pass through any other integer points in between (because $\gcd(n,n-1)=1$), then by Dumas' theorem the polynomial is irreducible.

For more information about these (and another) tools related to polynomial irreducibility, see this answer List of Generalizations of Common Questions. Also the polynomial with reverted coefficients is special case of this one: Irreducibility of $f(x)=x^n-p^m$ .

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Hint: I suggest you find the complex roots (not so hard), i.e., factorize the polynomial into a product of unary complex polynomials. Then show that it is impossible to partition these unary factors into two nonempty sets so that the product of each is a rational polynomial.

Focus on the constant term!

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