证明e的x-1次方>x的n次方/n!x属于1到正无穷

e^(x-1) > (x^n)/n!
n=1
L.S = e^(x-1)
R.S =x
consider
f(x) = e^(x-1) - x
f'(x) = e^(x-1) - 1 > 0 ( for x ∈（1,+∞)
f(x) > f(1) = 0
=> f(x) >0
e^(x-1) > x
for n=1 is true
Assum n = k is true, ie
e^(x-1) > (x^k)/k!
forn = (k+1)
LS = e^(x-1)
RS =(x^(k+1))/(k+1)!
consider
g(x)= e^(x-1) - (x^(k+1))/(k+1)!
g'(x) = e^(x-1) - (k+1)x^k/(k+1)!
= e^(x-1) - x^k/k!
> e^(x-1) - e^(x-1) =0
g(x)> g(1) = 1 - 1/(k+1)! > 0
g(x) > 0
=> e^(x-1) > (x^(k+1))/(k+1)!
By MI, it is true for all n ∈N