Summing up

$\begin{aligned} S&=\sum_{n=1}^{\infty} \dfrac{n^4}{(n+1)!}\\ &=\sum_{n=1}^{\infty} \dfrac{n^4-n^2+n^2-1+1}{(n+1)!}\\ &=\sum_{n=1}^{\infty} \dfrac{n^2(n+1)(n-1)}{(n+1)!}+\sum_{n=1}^{\infty}\dfrac{(n+1)(n-1)}{(n+1)!}+\sum_{n=1}^{\infty}\dfrac{1}{(n+1)!}\\ &=\sum_{\color{#3D99F6}n=1}^{\infty}\dfrac{n(n-1)}{(n-1)!}+\sum_{n=1}^{\infty}\dfrac{n-1}{n!}+e-2\\ &=\sum_{\color{#D61F06}n=2}^{\infty}\dfrac{n(n-1)}{(n-1)!}+\mathbin{\color{#3D99F6}\sum_{n=1}^{\infty}\left[\dfrac{1}{(n-1)!}-\dfrac{1}{n!}\right]}+e-2\\ &=\sum_{n=2}^{\infty}\left[\dfrac{n}{(n-2)!}\right]+\mathbin{\color{#3D99F6}1}+e-2\phantom{ccccccccccccccccccccc}\color{#3D99F6}\text{by Telescoping series sum}\\ &=\sum_{n=2}^{\infty}\left[\dfrac{n-2+2}{(n-2)!}\right]+e-1\\ &=\sum_{\color{#3D99F6}n=2}^{\infty}\dfrac{n-2}{(n-2)!}+\sum_{n=2}^{\infty}\dfrac{2}{(n-2)!}+e-1\\ &=\sum_{\color{#D61F06}n=3}^{\infty}\dfrac{n-2}{(n-2)!}+2e+e-1=\sum_{n=3}^{\infty}\dfrac{1}{(n-3)!}+3e-1=4e-1\approx\boxed{9.873} \end{aligned}$

$\begin{aligned} S & = \sum_{n=1}^\infty \frac {n^4}{(n+1)!} = \sum_{n=1}^\infty \frac {(n-1)^4}{n!} \\ & = \sum_{n=1}^\infty \frac {n^4-4n^3+ 6n^2-4n+1}{n!} \\ & = \sum_{n=1}^\infty \frac {n^4}{n!} - 4 \sum_{n=1}^\infty \frac {n^3}{n!} + 6 \sum_{n=1}^\infty \frac {n^2}{n!} - 4 \sum_{n=1}^\infty \frac n{n!} + \sum_{n=1}^\infty \frac 1{n!} \\ & = 15e - 4(5e) + 6(2e) - 4e + e - 1 & \small \color{#3D99F6} \text{See note.} \\ & = 4e -1 \\ & \approx \boxed{9.873} \end{aligned}$

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Note:

$\begin{aligned} \sum_{n=1}^\infty \frac 1{n!} & = \sum_{n=0}^\infty \frac 1{n!} - 1 = e - 1 \\ \sum_{n=1}^\infty \frac n{n!} & = \sum_{n=1}^\infty \frac 1{(n-1)!} = \sum_{n=0}^\infty \frac 1{n!} = e \\ \sum_{n=1}^\infty \frac {n^2}{n!} & = \sum_{n=1}^\infty \frac n{(n-1)!} = \sum_{n=0}^\infty \frac {n+1}{n!} = \sum_{n=0}^\infty \frac n{n!} + \sum_{n=0}^\infty \frac 1{n!} = \sum_{n=1}^\infty \frac n{n!} + e = 2e \\ \sum_{n=1}^\infty \frac {n^3}{n!} & = \sum_{n=1}^\infty \frac {n^2}{(n-1)!} = \sum_{n=0}^\infty \frac {(n+1)^2}{n!} = \sum_{n=0}^\infty \frac {n^2+2n+1}{n!} = 2e+2e+e = 5e \\ \sum_{n=1}^\infty \frac {n^4}{n!} & = \sum_{n=1}^\infty \frac {n^3}{(n-1)!} = \sum_{n=0}^\infty \frac {(n+1)^3}{n!} = \sum_{n=0}^\infty \frac {n^3+3n^2+3n+1}{n!} = 5e+6e+3e+e = 15e \end{aligned}$

$S=\small\displaystyle\sum_{n=1}^{\infty}\frac{n^4}{(n+1)!}$

$=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{(n^2+1)(n^2-1)}{(n+1)!} +\frac{1}{(n+1)!}\right]$ $=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{(n^2+1)(n-1)}{n!} +\frac{1}{(n+1)!}\right]$ $=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{n^3-n^2+n-1}{n!} +\frac{1}{(n+1)!}\right]$

$=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{(n^3-n^2)}{n!} +{\color{#3D99F6}\frac{1}{(n-1)!} - \frac{1}{n!}} +\frac{1}{(n+1)!}\right]$

$=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{n(n-1)}{(n-1)!} +{\color{#3D99F6}\frac{1}{(n-1)!} - \frac{1}{n!}} +\frac{1}{(n+1)!}\right]$

$=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{n}{(n-2)!} +{\color{#3D99F6}\frac{1}{(n-1)!} - \frac{1}{n!} }+\frac{1}{(n+1)!}\right]$

$=\small\displaystyle\sum_{n=1}^{\infty}\left[\frac{1}{(n-3)!} +\frac{2}{(n-2)!}+{\color{#3D99F6}\frac{1}{(n-1)!} - \frac{1}{n!}}+\frac{1}{(n+1)!}\right]$

$=\small e + 2e +{\color{#3D99F6}1} + (e-2) = 4e -1 = \boxed{9.873}$