Although the following does not provide another proof (perhaps it is possiblypossible to attempt one on this basis) I found it nice to see the following pictures.
Let's take from the series $f(x) = \sum_{k=0}^\infty {x^k \over \sqrt{k!}}$ the following variants in the same spirit as we have the hyperbolic and trigonometric series from the exponential-series:
$$\begin{array}{}
\exp_{\tiny \sqrt{\,}}(x) &=& f(x) \\
\cosh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k} \over \sqrt{(2k)!}} \\
\sinh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\tanh_{\tiny \sqrt{\,}}(x) &=& { \sinh_{\tiny \sqrt{\,}}(x)\over \cosh_{\tiny \sqrt{\,}}(x) } \\
\cos_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k} \over \sqrt{(2k)!}} \\
\sin_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\end{array}$$$$\begin{array}{}
\small \exp_{\tiny \sqrt{\,}}(x) &=& f(x) \\
\small \cosh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k} \over \sqrt{(2k)!}} \\
\small \sinh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\small \tanh_{\tiny \sqrt{\,}}(x) &=& { \sinh_{\tiny \sqrt{\,}}(x)\over \cosh_{\tiny \sqrt{\,}}(x) } \\
\small \cos_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k} \over \sqrt{(2k)!}} \\
\small \sin_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (-1)^k {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\end{array}$$
The answer to your question is equivalent to say, that always (="for real $x$")
- $\small \cosh_{\tiny \sqrt{\,}}(x)$ is larger than $\small \sinh_{\tiny \sqrt{\,}}(x) $ $\qquad \qquad$ or that
- $\small \mid \tanh_{\tiny \sqrt{\,}}(x) \mid \lt 1$
To illustrate this I've plotted the $\sinh_{\tiny \sqrt{\,}}$ and $\cosh_{\tiny \sqrt{\,}}$-curves:
![bild1][1]
This gives surely an extremely familiar impression...
The $\tanh_{\tiny \sqrt{\,}}$-curve looks completely familiar too: ![bild2][2]
and the image suggests, that indeed the absolute value of $\small \tanh_{\tiny \sqrt{\,}}(x) $ very likely is smaller than $1$ for all real $x$.
However, things are different for the $\sin_{\tiny \sqrt{\,}}$ and $\cos_{\tiny \sqrt{\,}}$ curves - they deviate strongly from the nicely periodic common trigonometric functions:
![bild3][3]
and combined they do not give a circle, but some ugly thing, strongly distorted (y-axis by $\small \cos_{\tiny \sqrt{ \,} }(\phi)$, x-axis by $\small \sin_{\tiny \sqrt{ \,} }(\phi)$, $\phi$ from $-5$ to $+5$) :
![bild4][4]
[1]: https://i.sstatic.net/zKZ6I.png [2]: https://i.sstatic.net/DT4At.png [3]: https://i.sstatic.net/yiYpF.png [4]: https://i.sstatic.net/YZSCk.png