how compute $\int_0^1\frac{(1-t)^n}{n!}e^tdt$ I'm trying to compute
$$\int_0^1\frac{(1-t)^n}{n!}e^tdt$$
but excepted making a $n$ integration by part, I don't see how can I do it. So, is there an easier way ?
 A: Another way
Following your first idea, let $$I_n=\int_0^1\frac{(1-t)^n}{n!}e^t\mathrm d t.$$
By an integration by part, you can get that $$I_{n}=\frac{e}{n!}+I_{n-1},$$
for all $n$. Using induction, the result follow. 
A: $$R(n)=\int_{0}^{1}\frac{(1-t)^n}{n!}e^t\,dt $$
can be seen as an integral remainder for a Taylor series. In particular, we have:
$$\begin{eqnarray*} R(n) = \frac{1}{n!}\int_{0}^{1} t^n e^{1-t}\,dt &=& \frac{e}{n!}\left.\left(t^n+nt^{n+1}+n(n-1)t^{n-2}+\ldots\right)e^{-t}\right|_{0}^{1}\\&=&e-\sum_{k=0}^{n}\frac{1}{k!}=\sum_{k>n}\frac{1}{k!}. \end{eqnarray*}$$
Since $1\leq e^t\leq e$ for $t\in[0,1]$, we have $R(n)=\frac{C}{(n+1)!}$ with $1\leq C\leq e$.
A: We have $$\frac{1}{n!}\int_{0}^{1}\left(1-t\right)^{n}e^{t}dt=\frac{1}{n!}\sum_{k\geq0}\frac{1}{k!}\int_{0}^{1}\left(1-t\right)^{n}t^{k}dt
 $$ $$\frac{1}{n!}\sum_{k\geq0}\frac{B\left(n+1,k+1\right)}{k!}=\sum_{k\geq0}\frac{1}{\left(n+k+1\right)!}=\color{red}{e-\sum_{k=0}^{n}\frac{1}{k!}}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Since the 'integrand structure' reminds us the exponential series, it's interesting to explore the use of a Generating Function. Namely,

\begin{align}
\color{#f00}{\int_{0}^{1}{\pars{1 - t}^{n} \over n!}\,\expo{t}\,\dd t} & =
\bracks{z^{n}}\bracks{\sum_{n = 0}^{\infty}z^{n}
\int_{0}^{1}{\pars{1 - t}^{n} \over n!}\,\expo{t}\,\dd t}
\\[5mm] & =
\bracks{z^{n}}\pars{\int_{0}^{1}\braces{\sum_{n = 0}^{\infty}
{\bracks{z\pars{1 - t}}^{n} \over n!}}\,\expo{t}\,\dd t}
\\[5mm] & =
\bracks{z^{n}}\bracks{\int_{0}^{1}\expo{z\pars{1 - t}}\expo{t}\,\dd t} =
\bracks{z^{n}}\bracks{\expo{z}\int_{0}^{1}\expo{\pars{1 - z}t}\,\dd t}
\\[5mm] & =
\bracks{z^{n}}{\expo{} - \expo{z}\over 1 - z} =
\bracks{z^{n}}\sum_{n = 0}^{\infty}\expo{}z^{n} -
\bracks{z^{n}}\sum_{k = 0}^{\infty}{z^{k} \over k!}
\sum_{n = 0}^{\infty}z^{n}
\\[5mm] & =
\expo{} - \bracks{z^{n}}\sum_{k = 0}^{\infty}{z^{k} \over k!}
\sum_{n = k}^{\infty}z^{n - k} =
\expo{} - \bracks{z^{n}}\sum_{n = 0}^{\infty}z^{n}
\sum_{k = 0}^{n}{1 \over k!}
\\[5mm] & =
\color{#f00}{\expo{} - \sum_{k = 0}^{n}{1 \over k!}}
\end{align}
