Binomial coefficients: how to prove an inequality on the $p$-adic valuation?

Since $\nu_p(n!)=\sum_i\lfloor n/p^i\rfloor$, $\nu_p\binom nm=\sum_i(\lfloor n/p^i\rfloor-\lfloor m/p^i\rfloor-\lfloor (n-m)/p^i\rfloor)$. Observe that each summand is $\le 1$, and for $i>\left\lfloor\frac{\ln n}{\ln p}\right\rfloor$ it's clearly $0$. That gives $v_{p}\binom{n}{m}\le \left\lfloor\frac{\ln n}{\ln p}\right\rfloor$.

Finally observe that for $i\le\nu_p(m)$ $\lfloor m/p^i\rfloor=m/p^i$, so ($\lfloor m/p^i\rfloor+\lfloor x/p^i\rfloor=\lfloor(m+x)/p^i\rfloor$ and) the corresponding summand is also $0$. That gives the inequality in question.

Theorem (Kummer): $\nu_p {n \choose m}$ is the number of carries it takes to add $m$ and $n-m$ in base $p$.

Now note that $\lfloor \frac{\ln n}{\ln p} \rfloor$ is the maximum possible number of carries and the last $\nu_p(m)$ digits of $m$ cannot be associated to any carries.

Kummer's theorem itself is not hard to prove; it more or less follows from the last identity you list.