How to approximate $\sum_{k=1}^n k!$ using Stirling's formula? How to find summation of the first $n$ factorials, 
$$1! + 2! + \cdots + n!$$
I know there's no direct formula, but how can it be estimated using Stirling's formula?
Another question :
Why can't we find the summation of n! ?
Why there's no direct formula?
 A: Stirling's formula gives us that $$n! \sim \sqrt{2 \pi n} \left( \dfrac{n}e\right)^n$$ i.e. $$\lim_{n \to \infty} \dfrac{n!}{\sqrt{2 \pi n} \left( \dfrac{n}e\right)^n} = 1$$
It is not hard to show that your sum, $$\sum_{k=1}^{n} k! \sim n!$$ and hence $$\sum_{k=1}^{n} k! \sim \sqrt{2 \pi n} \left( \dfrac{n}e\right)^n$$
EDIT To see that $\displaystyle \sum_{k=1}^{n} k! \sim n!$, note that
\begin{align}
\sum_{k=1}^{n} k! & = n! \left( 1 + \dfrac1n + \dfrac1{n(n-1)} + \dfrac1{n(n-1)(n-2)} + \cdots + \dfrac1{n!}\right)\\
& \leq n! \left( 1 + \dfrac1n + \dfrac{n-1}{n(n-1)}\right)\\
& = n! \left( 1 + \dfrac2n\right)
\end{align}
Hence, $\displaystyle \sum_{k=1}^{n} k! \sim n!$.
A: There is the direct formula:
$$\sum_{k=1}^{n-1} \Gamma(k)=(-1)^{n+1}\Gamma(n)(!(-n))+C$$
Where !(x) is subfactorial.
A: To obtain better approximations, note that for large $n$, we have
$$
\sum\limits_{k = 1}^n {k!}  = n!\left( {1 + \frac{1}{n} + \frac{1}{{n\left( {n - 1} \right)}} + \frac{1}{{n\left( {n - 1} \right)\left( {n - 2} \right)}} +  \cdots } \right) = n!\left( {1 + \frac{1}{n} + \frac{1}{{n^2 }} + \frac{2}{{n^3 }} +  \cdots } \right).
$$
Substituting Stirling's formula
$$
n! \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{1}{{12n}} + \frac{1}{{288n^2 }} - \frac{{139}}{{51840n^3 }} -  \cdots } \right)
$$
yields
$$
\sum\limits_{k = 1}^n {k!}  \sim \left( {\frac{n}{e}} \right)^n \sqrt {2\pi n} \left( {1 + \frac{{13}}{{12n}} + \frac{{313}}{{288n^2 }} + \frac{{108041}}{{51840n^3 }} +  \cdots } \right) .
$$
