Proof for the inequality $x_n>y_m$ for $n>m$, where $x_n = \left(1+\frac1n\right)^n$ and $y_n = \sum\limits_{k=0}^n \frac1{k!}$ I encountered the following inequality in A Basic Course in Real Analysis by Kumaresan and Ajit Kumar, page 45. I am unable to get a proof for this inequality.
Given,
$x_n = 1 + \sum\limits_{k=1}^{n} \frac{n!}{k! (n-k)!} n^{-k}$
$y_n = \sum\limits_{k=0}^{n} \frac{1}{k!}$ 
Prove that
If $n,m \in \mathbb{N} $ and $n > m$, then $x_n > y_m$
Edit: The inequality was taken as an obvious result as shown in this picture:
Portion from the textbook
 A: The way that $\left(1+\frac1n\right)^n$ is written, seems to indicate that the author wants us to compare the sums term by term.
$$
\begin{align}
\frac{n!}{k!(n-k)!}n^{-k}
&=\frac1{k!}\frac{n!}{n^k(n-k)!}\\
&=\frac1{k!}\prod_{j=0}^{k-1}\frac{n-j}n\\
&\le\frac1{k!}
\end{align}
$$
The inequality seems to be backward.

Indeed, since
$$
\begin{align}
\sum_{k=n+1}^\infty\frac1{k!}
&=\frac1{(n+1)!}\sum_{k=0}^\infty\frac1{(n+2)^{\overline{k}}}\\
&\le\frac1{(n+1)!}\sum_{k=0}^\infty\frac1{(n+2)^k}\\
&=\frac1{(n+1)!}\frac{n+2}{n+1}
\end{align}
$$
we have
$$
\bbox[5px,border:2px solid #C0A000]{y_n\ge e-\frac1{(n+1)!}\frac{n+2}{n+1}}
$$
Furthermore
$$
\begin{align}
n\log\left(1+\frac1n\right)
&=-n\log\left(1-\frac1{n+1}\right)\\
&=n\sum_{k=1}^\infty\frac1{k(n+1)^k}\\
&=\left(1-\frac1{n+1}\right)\sum_{k=1}^\infty\frac1{k(n+1)^{k-1}}\\
&=1-\sum_{k=1}^\infty\frac1{k(k+1)(n+1)^k}
\end{align}
$$
whereas
$$
\begin{align}
1+\log\left(\frac{2n+1}{2n+2}\right)
&=1+\log\left(1-\frac1{2n+2}\right)\\
&=1-\sum_{k=1}^\infty\frac1{k2^k(n+1)^k}
\end{align}
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{x_n\le e-\frac e{2n+2}}
$$
Therefore, we need $n\gt\frac e2(m+1)!\frac{m+1}{m+2}-1$ to get $x_n\gt y_m$.

For example, $x_n\gt y_5\implies n\ge841$ (formula above gives $838$), and $x_n\gt y_6\implies n\ge6006$ (formula above gives $5993$).
