the following inequality is true，but I can't prove it 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.

This is true for large $d$, and probably for all $d$. I'll prove that the sum is 
$$e^d(1-1/d+1/d^2+O(1/d^{2.5+\epsilon}))$$
and leave the explicit bounds to you. 
Set $k=d+\ell$. For $|\ell| \leq d^{0.5+\epsilon}$, we have
$$1-1/(2d+2-k) = 1-\frac{1}{d} \frac{1}{1-(\ell-2)/d} = 1-\frac{1}{d} - \frac{\ell-2}{d^2} - \frac{(\ell-2)^2}{d^3} + O(d^{-2.5+3 \epsilon})$$
$$=1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3} + O(d^{-2.5+\epsilon}).$$
We will show later that the difference between your sum and 
$$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{\ell}{d^2} + \frac{2 d - \ell^2}{d^3}  \right)$$
is very small, where $\ell = k-d$. Assuming that, let's compute the new sum. With the aid of Mathematica, 
$$\sum_{k=0}^{\infty} \frac{x^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3}  \right) = e^x \left( 1-\frac{1}{d}+\frac{x}{d^2}-\frac{x^2}{d^3} + \frac{2}{d^2} - \frac{x}{d^3} \right). $$
Plugging in $x=d$, 
$$\sum_{k=0}^{\infty} \frac{d^k}{k!} \left( 1-\frac{1}{d} - \frac{k-d}{d^2} + \frac{2 d - (k-d)^2}{d^3}  \right) = e^d \left( 1-\frac{1}{d} + \frac{1}{d^2} \right).$$
The error coming from $O(d^{-2.5+\epsilon}) \sum d^k/k!$ is $e^d O(d^{-2.5+\epsilon})$. 
We now just need to think about the error coming from discarding the terms with $|\ell|>d^{0.5+\epsilon}$. No matter what $\ell$ is, that error is no worse than $d^k/k! \cdot ( O(\ell^2/d^3) + O(1))$. But, for $|\ell| > d^{0.5 + \epsilon}$, we have $d^k/k! \leq e^{-d^{2 \epsilon}}$ as I pointed out on math.SE, so the contribution from these terms is exponentially small.
I have not found a slick way to give a proof for all $d$, rather than an asymptotic result. 
