用数学归纳法.
n=0时,
t^n-nt+(n-1)=1-0+（0-1）=0满足≥0.
假设对n=任意自然数k,t>0,满足t^k-kt+(k-1)≥0
n=k+1时,
t^(k+1)-(k+1)t+(k+1-1)
=t*t^k-k*t-t+k
=t*(t^k-kt+k-1) + k*t*t-2kt + k
=t*(t^k-kt+k-1) + k*(t^2-2t+1)
=t*(t^k-kt+k-1) +k*(t-1)^2
>=0 + k*(t-1)^2
>= 0