The inequality is
\begin{equation*} 
\sum_{k=1}^{2d}\left(1-\frac{1}{2d+2-k}\right)\frac{d^k}{k!}>e^d\left(1-\frac{1}{d}\right)
\end{equation*}
for all integer $d\geq 1$. I use computer to verify it for $d\leq 50$, and find it is true, but I can't prove it.  Thanks for your answer.
[Edited mostly to fix a typo noted by David Speyer]

The following analysis simplifies and completes the
"routine but somewhat unpleasant" task of
recovering the actual inequality from the asymptotic analysis.

The idea is that once we've obtained the asymptotic expansion
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!}
\sim e^d \left( 1 - \frac1{d} + \frac1{d^2} - \frac2{d^3} \cdots \right)
$$
by expanding $1 - 1/(2d+2-k)$ in a power series about $k=d$,
we should be able to replace $1 - 1/(2d+2-k)$ by something smaller
that can be summed exactly and is close enough that the result
is within a small enough multiple of $e^d$ to maintain the
desired inequality.

Because it takes about $2m$ terms of the power series in $k$
to get within $O(1/d^m)$, I had to match the power series to within $O(k-d)^6$.
Let
$$
A_6(k) =
\frac{d+1}{d+2}
- \frac{k-d}{(d+2)^2}
- \frac{(k-d)^2}{(d+2)^3}
- \frac{(k-d)^3}{(d+2)^4}
- \frac{(k-d)^4}{(d+2)^5}
- \frac{(k-d)^5}{(d+2)^6}
- \frac{(k-d)^6}{2(d+2)^6},
$$
so the final term has denominator $2(d+2)^6$ instead of $(d+2)^7$.
Then
$$
1 - \frac1{2d+2-k}
 = A_6(k) + \frac{(k-d)^6(2d-k)}{2(d+2)^6(2d+2-k)} \geq A_6(k)
$$
for all $k \leq 2d$.  Hence
$$
\sum_{k=1}^{2d} \left( 1 - \frac1{2d+2-k} \right) \frac{d^k}{k!} >
\sum_{k=1}^{2d} A_6(k) \frac{d^k}{k!}.
$$
On the other hand, since $A_6(k)$ is a polynomial in $k$,
the power series $\sum_{k=0}^\infty A_6(k) d^k/k!$ is elementary
(see my earlier answer for the explanation; David Speyer implicitly
used this too in the calculation "with the aid of Mathematica").
I find
$$
\sum_{k=0}^\infty A_6(k) \frac{d^k}{k!}
= \frac{2d^6 + 22d^5 + 98d^4 + 102d^3 + 229d^2 + 193d + 64}{2(d+2)^6} e^d
$$ $$
= \left(
 1 - \frac1d + \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6}
\right) \cdot e^d > \left(1 - \frac1d\right) e^d.
$$
We're not quite finished, because we need a lower bound on
$\sum_{k=1}^{2d} A_6(k) d^k/k!$, not $\sum_{k=0}^\infty$.
However, once $d$ is at all large the terms with $k=0$ and $k>2d$
are negligible compared with our lower bound
$$
\sum_{k=0}^\infty A_6(k) d^k/k! - \left(1 - \frac1d\right) e^d
\geq \frac{2d^5 + 5d^4 + 69d^3 + 289d^2 + 320d + 128}{2d(d+2)^6} e^d >
\frac{d^4}{(d+2)^6} e^d.
$$
Indeed the $k=0$ term is less than $1$, and for $k>2d$ we have
$A_6(k) < A_6(2d) = 1/2$ while $d^k/k!$ is exponentially smaller than $e^d$:
$$
\sum_{k=2d+1}^\infty \frac{d^k}{k!} <
 2^{-2d} \sum_{k=2d+1}^\infty \frac{(2d)^k}{k!} <
 2^{-2d} \sum_{k=0}^\infty \frac{(2d)^k}{k!} = (e/2)^{2d}.
$$
So we're done once
$$
1 + \frac12 \left(\frac{e}{2}\right)^{2d} < \frac{d^4}{(d+2)^6} e^d,
$$
which happens once $d \geq 14$.  Since the desired inequality
has already been verified numerically up to $d=50$, we're done. QED
