This can't be that Short!

First we prove $\sum_{i=a}^b i*i! = (b+1)! - a!$

Proof:- $\sum_{i=a}^b i*i!$

$= \sum_{i=a}^b (i+1)*i! - i!$

$= \sum_{i=a}^b (i+1)! - i!$

$= (a+1)! - a! + (a+2)! - (a+1)! + \dots + (b+1)! - b!$

$= (b+1)! - a!$ .

Now to the question.

$\sum_{k=1}^m (k^2 + 1)k!$

$=\sum_{k=1}^m (k^2 +2k+ 1)k! - 2k*k!$

$=\sum_{k=1}^m (k+1)^2k! - 2\sum_{k=1}^m k*k!$

$= \sum_{k=1}^m (k+1)*(k+1)! - 2((m+1)! - 1!)$

Putting $l = k+1$ , $l =2$ when $k = 1$ and $l = m+1$ when $k = m$ .

$= \sum_{l=0}^{m+1}l*l! - 2(m+1)! + 2$

$= (m+2)! - 2! - 2(m+1)! + 2$

$= (m+1)!(m+2 - 2)$

$= m*(m+1)!$

Comparing with $1999*(2000)!$ , we see that $m = \boxed{1999}$

We know $r*r! = [(r+1)-r]r!= (r+1)!-r!$ .

So, $(k^2+1)k!= k(k*k!)+k!$

$=k[(k+1)!-k!]+k!$

$=[(k+2)-2](k+1)!-k*k!+k!$

$=(k+2)!-2(k+1)!-(k+1)!+2k!$

$=[(k+2)!-(k+1)!]-2[(k+1)!-k!]$

Now, $\sum^m_{k=1}[(k^2+1)k!]$

$= \sum^m_{k=1}[(k+2)!-(k+1)!]- 2\sum^m_{k=1}[(k+1)!-k!]$

$=(m+2)!-2!-2[(m+1)!-1]$ $= (m+1)![m+2-1]=m*(m+1)!=1999*(2000)!$

So, $m=\boxed{1999}$ .

$\left( { k }^{ 2 }+1 \right) k!=\left\{ \left( k+2 \right) \left( k+1 \right) -3\left( k+1 \right) +2 \right\} k!\\ \\ \quad =\left( k+2 \right) !-3\left( k+1 \right) !+2k!\\ \\ \quad =\left\{ \left( k+2 \right) !-\left( k+1 \right) ! \right\} -2\left\{ \left( k+1 \right) !-k! \right\} \\ \\ telescoping...\\ \\ \sum _{ k=1 }^{ m }{ \left( { k }^{ 2 }+1 \right) k!=\left\{ \left( m+2 \right) !-2! \right\} -2\left\{ \left( m+1 \right) !-1! \right\} } \\ \\ =\left\{ \left( m+2 \right) !-2\left( m+1 \right) ! \right\} -\left\{ 2!-2.1! \right\} \\ \\ =m\left( m+1 \right) !=1999\times 2000!\\ \\ \Rightarrow m=1999$

Note: The following solution is hinted by an online friend FAlin from Taiwan's forum telnet://ptt.cc or https://www.ptt.cc/bbs/Math/

Note that $(k^2+1)k! = (k^2+k-k+1)k! = [k(k+1)-(k-1)]k! = k(k+1)!-(k-1)k!$ then

$\sum_{k=1}^m [(k^2+1)k!] = \sum_{k=1}^m [k(k+1)!-(k-1)k!] = \sum_{k=1}^m [-(k-1)k!+k(k+1)!]$

$= [-0*1!+1*2!]+[-1*2!+2*3!]+[-2*3!+3*4!]+[-3*4!+4*5!]+[-4*5!+5*6!]+...$ (telescoping)

$+[-(m-1)m!+m(m+1)!] = m(m+1)!$

Therefore, 1999*2000! = $\sum_{k=1}^{1999} [(k^2+1)k!]$ , m=1999

$(k^{2}+1)(k!)$ is: For $k_1=(1+1)(1!)=2$

For $k_2=(4+1)(2!)=10$

For $k_3=(9+1)(3!)=60$

For $k_4=(16+1)(4!)=408$

For $k_5=(25+1)(5!)=3120$

Let $Sk_i=\sum\limits_{k=1}^{k_i}[(k^2+1)(k!)]$

$Sk_1=2 \Rightarrow 2!\times1$

$Sk_2=12 \Rightarrow 3!\times2$

$Sk_3=72 \Rightarrow 4!\times3$

$Sk_4=480 \Rightarrow 5!\times4$

By induction proof that: $\sum\limits_{k=1}^{m}=(m)(m+1)!$

Step 1 $\sum\limits_{k=1}^{1}[(k^2+1)(k!)]=(1)(1+1)!=2$ $k_1=(1+1)(1!)=2$

Step 2 $\sum\limits_{k=1}^{m}=(m)(m+1)!$ is true.

Step 3 Proof that $\sum\limits_{k=1}^{m+1}=(m+1)(m+2)!$ $\sum\limits_{k=1}^{m+1}[(k^2+1)(k!)]=(m)(m+1)!+((m+1)^2+1)(m+1)!$ $\sum\limits_{k=1}^{m+1}[(k^2+1)(k!)]=(m)(m+1)!+((m^2+2m+1+1)(m+1)!$ $\sum\limits_{k=1}^{m+1}[(k^2+1)(k!)]=(m+1)!(m^2+3m+2)$ $\sum\limits_{k=1}^{m+1}[(k^2+1)(k!)]=(m+1)!(m+2)(m+1)$ $\sum\limits_{k=1}^{m+1}[(k^2+1)(k!)]=(m+2)!(m+1)\blacksquare$

For $Sk_i=\sum\limits_{k=1}^{m}[(k^2+1)(k!)=1999\times 2000!$ $\boxed{m=1999}$