Let $p$ be a prime, and consider $$S\_p(a)=\sum\_{\substack{1\le j\le a-1\\(p-1)\mid j}}\binom{a}{j}\;.$$
I have a rather complicated (15 lines) proof that $S\_p(a)\equiv0\pmod{p}$. This must be
extremely classical: is there a simple direct proof ?



Analogue of Fermat's "little" theorem
Let $$P(x)=(1+x)^a-1-x^a=\sum\_{1 \le j \le a-1} \binom{a}{j}x^j.$$ Working in a field $F$ where $|\{\mu \in F: \mu^{p-1}=1\}|=p-1$ (roots of unity of order $p-1$ exist), we have
$$ \frac{1}{p-1}\sum\_{\mu^{p-1}=1}P(x \mu) = \sum\_{\substack{1 \le j \le a-1\\(p-1)\mid j}} \binom{a}{j}x^j.$$
We now specialize the field to $\mathbb{F}\_p$, and let $x=1$:
$$ -\sum\_{\mu \in \mathbb{F}\_p^{\times}}P( \mu) = \sum\_{\substack{1 \le j \le a-1\\(p-1)\mid j}} \binom{a}{j}.$$
To conclude, observe that $P(0)=0$ and $|\mathbb{F}\_p| = p$, so $$\sum\_{\mu \in \mathbb{F}\_p^{\times}} P(\mu) = \sum\_{\mu \in \mathbb{F}\_p} P(\mu) = \sum\_{\mu \in \mathbb{F}\_p} ((1+\mu)^a - \mu^a)=0,$$
because $\mu\mapsto \mu+1$ is a permutation of $\mathbb{F}\_p$.




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A slightly more general congruence is due to Glaisher (1899), as I've found in a survey by Granville, see [equation (11) here](https://dms.umontreal.ca/%7Eandrew/Binomial/sumsofbcs.html). The precise reference is Glaisher, J. W. L., "A congruence theorem relating to sums of binomial-theorem coefficients.", Quart. J. 30, 150-156, 349-360, 361-383 (1899). See [here for the zbMath review](https://zbmath.org/?q=an%3A29.0152.05) in German. There is no entry in MathSciNet.


Namely, Glaisher proved that for any $0 \le j\_0 \le p-1$ we have
$$\sum\_{\substack{1 \le j \le a \\ j \equiv j\_0 \bmod (p-1)}} \binom{a}{j} \equiv \binom{k}{j\_0} \bmod p$$
where $k$ is any positive integer with $k \equiv a \bmod p$ Applying this with $j\_0=0$ and $k=a$, and observing that the term $j=a$ contributes $1 \bmod p$, your result follows. Clicking on equation (11) in the link above leads to a [detailed proof](https://dms.umontreal.ca/%7Eandrew/Binomial/pfeq11.html) which utilizes Lucas' theorem.


