Tail bound for Poisson random variable Is the following fact about Poisson random variables true?

For any $\lambda \in (0,1)$ and integer $k > 0$, if $X$ is a Poisson random variable with mean $k \lambda$, then $\Pr(X < k) \geq e^{-\lambda}$.

It clearly holds for $k=1$, and for any fixed $\lambda$ it's easy to see that it holds for all sufficiently large $k$, but for intermediate values of $k$ it's not obvious to me.
 A: The argument below is a bit weird because it is a mixture of an explicit computation and a general handwaving (rigorous handwaving, of course), but, when figuring it out at 70 mph
under medium strength rain, I could use neither pen and paper, nor the full power of my imagination, so I just took whatever came easily from both worlds and made this crazy hybrid.
Step 1. Since the gaps between Poisson are governed by independent exponential distributions,
you are asking to prove that $P(Z_k=\sum_1^k Y_k>k)\ge e^{-\lambda}$ where $Y_k\in\exp(\lambda)$ are independent.
Step 2. The density of $Z_k$ is proportional to $t^ke^{-\lambda t}$. Thus, it is always skewed to the right around $k$ in the sense that if the probability to be on the right of $k$ is $a$, and the excess $Z_k-k$ conditioned upon $Z_k>k$ is $V$, then $Z_k$ can be coupled with the random variable $W_k$ that is the mixture of $V$ and $-V$ with probabilities $a$ and $1-a$ correspondingly so that $Z_k\ge W_k$. To see it, it suffices to show that the density of $V$ and the density of $k-Z_k$ conditioned upon $Z_k<k$ cross at only one point (then the crossing is of the right type because the density of $V$ survives at infinity but that of $k-Z_k$ dies beyond $k$). This is equivalent to showing that $(k+s)^k e^{-\lambda s}=c(k-s)^k e^{\lambda s}$ can have at most one solution on $(0,k)$, i.e., $\log(1+t)-\log(1-t)-2\lambda t$ is one-to-one on $(0,1)$. But it is a Taylor series with positive coefficients over odd powers of $t$ as long as the coefficient $2-2\lambda$ at $t$ is non-negative, i.e., exactly in the range of $\lambda$ you need.
Step 3. Induction on $k$. Suppose that we have two independent random variables $Z,Z'$ with continuous densities skewed to the right about $k,k'$ in the above sense. Let $a=P(Z>k)$, $a'=P(Z'>k')$. We want to show that we then have $P(Z+Z'>k+k')\ge\min(a,a')$. WLOG, $a\le a'$. Clearly, we are worse off with $W$ and $W'$. But then the probability is
$$
\begin{aligned}
&aa'+a(1-a')P(V>V')+a'(1-a)P(V'>V)
\cr
&\ge aa'+a(1-a')[P(V>V')+P(V'>V)]=a\,.
\end{aligned}
$$
A: This is not a full answer, but
according to this explanation
I think the question asks whether $\Gamma(k, k \lambda) \geq e^{-\lambda} \Gamma(k)$ when $k \in Z^+$, and $\lambda \in (0, 1)$.
Edit:
Some answers to this math.se question
have lower bounds for the upper incomplete gamma function.
A: A classical inequality of Teicher (1955) asserts

Proposition (Teicher). Let $X \sim \mathrm{Pois}(\lambda)$. Then, $\mathbb P(X \leq [\lambda]) > e^{-1}$.

A modification of his argument will allow us to prove the following.
Proposition. For $\lambda \in (0,1)$, let $X_{n\lambda} \sim \mathrm{Pois}(n\lambda)$. Then the sequence $b_n := b_{n,n\lambda} = \mathbb P(X_{n\lambda} < n)$ is monotonically increasing. In particular, $b_n \geq e^{-\lambda}$ for all $n$.
Proof. First, note that, by considering a Poisson process with rate 1, we have, 
$$
b_{n+1,\mu} = \sum_{x=0}^n \frac{e^{-\mu} \mu^x}{x!} = \int_\mu^\infty \frac{x^n e^{-x}}{n!} \,\mathrm{d}x \,,
$$
for all $n$ and $\mu$. Now,
$$
b_{n+1,n\lambda} - b_{n+1,(n+1)\lambda} = \int_{n\lambda}^{(n+1)\lambda} \frac{x^n e^{-x}}{n!} \,\mathrm{d}x = \int_0^1 (\lambda(y+n))^n \frac{e^{-\lambda(y+n)}}{n!} \lambda \,\mathrm{d}y \,,
$$
where the last equality follows from the substitution $y = (x-n\lambda)/\lambda$.
We can rewrite the last integral as
$$
\frac{e^{-\lambda n}(\lambda n)^n}{n!} \lambda \int_0^1 (1+y/n)^n e^{-\lambda y} \,\mathrm{dy} <  \frac{e^{-\lambda n}(\lambda n)^n}{n!} = b_{n+1,n\lambda} - b_{n,n\lambda} \,
$$
where the inequality follows from facts that $(1+y/n)^n < e^y$ and (upon integrating) $e^{1-\lambda} < \lambda^{-1}$, true for any $\lambda \in (0,1)$.
But, then $b_{n,n\lambda} < b_{n+1,(n+1)\lambda}$ which is what was to be shown. Since $b_{1,\lambda} = e^{-\lambda}$, the second part of the proposition statement holds.
A: Try $k = 2$ and $\lambda=2$.   Then
$P(X < k) = P(X = 0 ) + P(X = 1 ) = \exp(-4) + 4 \exp(-4) \approx  0.0915782 $
and
$\exp(-\lambda) = \exp(-2) \approx 0.135335.$
The hypothesis fails.
