First, observe that

$${m!\over(m-i)!} \le m^i $$

for $0\le i\le m$. (The two sides are equal for $i=0$ and $1$. Otherwise the inequality is strict: The left hand side is a product of $i$ positive integers, none greater than $m$.) It follows that, for $0\le i \le d \le m$, we have

$$\left({d\over m}\right)^d{m\choose i}= {d^i\over i!}\left({d\over m}\right)^{d-i}{m!\over m^i(m-i)!} \le {d^i\over i!}, $$

hence

$$\left({d\over m}\right)^d\sum_{i=0}^d {m\choose i} \le \sum_{i=0}^d {d^i\over i!} \lt \sum_{i=0}^\infty {d^i\over i!} =e^d. $$

The desired inequality follows.

(Note: The OP did not explicitly assume that $d\le m$, but it's reasonable to assume he or she meant to. In particular, if $d\gt em$, the OP's inequality is false.)