Answer: \(I_n=nI_{n-1}-e^{-1}.\)

For every positive integer \(n\), \(I_n\lt n!\).

Let

\(I_n=\int_0^1x^ne^{-x}\,\mathrm dx.\)

Integrate by parts with \(u=x^n\) and \(\mathrm dv=e^{-x}\,\mathrm dx\). Then

\(\mathrm du=nx^{n-1}\,\mathrm dx,\qquad v=-e^{-x}.\)

So

\(I_n=\left[-x^ne^{-x}\right]_0^1+n\int_0^1x^{n-1}e^{-x}\,\mathrm dx.\)

The boundary term is \(-e^{-1}\), hence

\(I_n=nI_{n-1}-e^{-1}.\)

Now prove \(I_n\lt n!\) by induction.

For \(n=1\),

\(I_1=1\cdot I_0-e^{-1}.\)

Since \(I_0=\int_0^1e^{-x}\,\mathrm dx=1-e^{-1}\),

\(I_1=1-2e^{-1}\lt 1=1!.\)

Assume \(I_k\lt k!\). Then

\(I_{k+1}=(k+1)I_k-e^{-1}\lt (k+1)k!-e^{-1}\lt (k+1)!.\)

Therefore, by induction,

\(I_n\lt n!\quad\text{for every positive integer }n.\)