Prove inductively that log(x) < x for x>0

Base: log1=0<1

Assume log(k)<k

Show log(k+1)<k+1

log(k+1) = log(k) + log(1/k + 1)

substitute k for log(k)

Show k + log(1/k + 1) < k + 1

log(1/k + 1) <1

log (1/k + 1) = log ((k+1)/k) = log(k+1) - log k

log (k+1) -log(k) < 1

log (k+1) < 1 + log (k)

Now what? I think I'm taking the wrong approach, but I can't see another one...