Congruence between Bernoulli numbers I fell on the following fact as a by-product:

If p is a prime number, then the sum of the Bernoulli numbers, from index 0 to p - 2, is congruent with - 1 modulo p.

Do you know a simple proof ? Or a reference to the literature ? Thanks beforehand.
Edit. It was not a very smart question. The congruence is an easy consequence of Bernoulli's formula for the sum of the $(p-2)$-powers of the natural numbers from 1 to $p-2$.
Edit. As suggested by barto, I give the proof. For $p = 2$, it is trivial, thus assume $p \geq 3$. Bernoulli's summation formula is (for $m \geq 1$) :
$$
(m+1) \sum_{k=1}^{n-1} k^m = \sum_{k=0}^m \binom{m + 1}{k} B_k n^{m + 1 - k}
$$
or in more detail
$$
(m+1)(1^m + 2^m + \cdots + (n-1)^m) = \sum_{k=0}^m \binom{m + 1}{k} B_k n^{m + 1 - k}.
$$
(I use the definition with $B_1 = -1/2$. The Wikipedia article :
https://en.wikipedia.org/wiki/Bernoulli_number#Sum_of_powers
uses the other definition, hence a slight difference.)
With $n = p-1$ and $m = p-2$, this becomes
(1) $(p-1) \sum_{k=1}^{p-2} k^{p-2} = \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k}$,
where the involved Bernoulli numbers are $p$-integral rational numbers.
Now, by Fermat's little theorem, $\sum_{k=1}^{p-1} k^{p-2}$ (with one more term than in the preceding sum) is modulo $p$ the sum of the inverses of the $p-1$ nonzero elements. Since inversion is a bijection, this sum is equal modulo $p$ to $\sum_{k=1}^{p-1} k = p(p-1)/2$, which is zero modulo $p$. Thus (1) gives
$-(p-1)(p-1)^{p-2} \equiv \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k} \pmod{p}$
$-1 \equiv \sum_{k=0}^{p-2} \binom{p-1}{k} B_k (p-1)^{p-1-k}\pmod{p}$.
But $\binom{p-1}{k} \equiv (-1)^{k} \pmod{p}$, thus the right member is congruent modulo $p$ to $\sum_{k=0}^{p-2} B_k$, and we are done.
Edit : here is a simpler proof. It is well known that, for $m \geq 2$,
$\sum_{i=0}^{m-1} \binom{m}{i} B_{i} = 0.$ Putting $m=p-1$, we find
$\sum_{i=0}^{p-2} \binom{p-1}{i} B_{i} = 0.$
Since $\binom{p-1}{k} \equiv (-1)^{k} \pmod{p}$ and the involved Bernoulli numbers are $p$-integral rational numbers, this gives
$\sum_{i=0}^{p-2} (-1)^{i} B_{i} \equiv 0 \pmod{p}.$
Now, $B_{1} = -1/2$ and for $i$ odd and $&gt; 1$, $B_{i} = 0$, hence the statement.
What you have is incorrect. $\displaystyle \sum_{k=0}^{p-2} B_k$ is not guaranteed to be an integer in the first place, let alone being $-1 \pmod{p}$. For instance, $$\sum_{k=0}^{3-2} B_k = \dfrac12$$
$$\sum_{k=0}^{5-2} B_k = \dfrac23$$
$$\sum_{k=0}^{7-2} B_k = \dfrac{19}{30}$$
What you have is incorrect. $\displaystyle \sum_{k=0}^{p-2} B_k$ is not guaranteed to be an integer in the first place, let alone being $-1 \pmod{p}$. For instance, $$\sum_{k=0}^{3-2} B_k = \dfrac12$$
$$\sum_{k=0}^{5-2} B_k = \dfrac23$$
$$\sum_{k=0}^{7-2} B_k = \dfrac{19}{30}$$