  In this paper we establish some new supercongruences motivated by the
well-known fact $\lim_{n\to\infty}(1+1/n)^n=e$. Let $p>3$ be a prime. We prove
that $$\sum_{k=0}^{p-1}\binom{-1/(p+1)}k^{p+1}\equiv 0\ \pmod{p^5}\ \ \
\mbox{and}\ \ \ \sum_{k=0}^{p-1}\binom{1/(p-1)}k^{p-1}\equiv
\frac{2}{3}p^4B_{p-3}\ \pmod{p^5},$$ where $B_0,B_1,B_2,\ldots$ are Bernoulli
numbers. We also show that for any $a\in\mathbb Z$ with $p\nmid a$ we have
$$\sum_{k=1}^{p-1}\frac1k\left(1+\frac ak\right)^k\equiv -1\pmod{p}\ \ \
\mbox{and}\ \ \ \sum_{k=1}^{p-1}\frac1{k^2}\left(1+\frac ak\right)^k\equiv
1+\frac 1{2a}\pmod{p}.$$
