Let $n$ and $1\leq k \leq n$ be natural numbers. Prove the inequality $$\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} \leq 1 - \frac{1}{e} $$
Equivalently, if $X\sim$ Bin($n$,$\frac{k}{n+1}$), prove that $\mathbb{P}[X\geq k] \leq 1 - \frac{1}{e}$.
My attempt: It may be helpful to show that the LHS tends to $1-\frac{1}{e}$ as $n \to \infty$ (already did that) and that the LHS is an increasing function on $n$ (have not done that).
HINT
Note that
$$\sum_{i=k}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} =\sum_{i=0}^n \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i} -\sum_{i=0}^{k-1} \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i}=1^n-\sum_{i=0}^{k-1} \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i}$$
and show that
$$\sum_{i=0}^{k-1} \binom{n}{i}\bigg(\frac{k}{n+1}\bigg)^i\bigg(1-\frac{k}{n+1}\bigg)^{n-i}\stackrel{k=1}\ge \sum_{i=0}^0 \binom{n}{i}\bigg(\frac{1}{n+1}\bigg)^i\bigg(1-\frac{1}{n+1}\bigg)^{n-i}=\bigg(1-\frac{1}{n+1}\bigg)^{n}\ge\frac 1e$$