I am reading a paper that references the result $$E_{p-1} \equiv 1 \mod p$$ where $p \geq 5$ is prime and $E_{p-1}$ is the Eisenstein series. But the reference references another text, which I have no access to.
Are there any other references to see why this is true? Or, is it obvious enough to post a proof here? Thanks either way.
The Eisenstein series is $$E_{p-1}=1-\frac{2(p-1)}{B_{p-1}}\sum_{n=1}^\infty\sigma_{p-2}(n)q^n$$ where the $B_k$ are Bernoulli numbers and $\sigma_k(n)=\sum_{d\mid n}d^k$. For prime $p\ge 5$, $p$ divides the denominator of $B_{p-1}$ by the von Staudt-Clausen congruences. So modulo $p$ every coefficient save the initial $1$ is a multiple of $p$ divided by a non-multiple of $p$.