# Underapproximating the exponential function from below [duplicate]

I think that for positive natural numbers t and n we have

$$\left(1+\frac{n}{t}\right)^t\ \le\ e^n\,.$$

1. Is this true? I have constructed a proof of it (which would probably take some lengthy typesetting here, and further I believe that the inequality could be very standard), but I wish to have an independent confirmation from the folks here.

2. If the answer to the previous question is "yes", is there a proof in the published, reviewed literature (book, article, thesis, etc.) that I can reference? We really don't want to show that we've reinvented the wheel.

EDIT: Since the answers confirmed the inequality, and there are even duplicate questions, I'm really interested in part 2 from now on.

## marked as duplicate by Martin R, Peter Foreman, callculus, Adrian Keister, Lee David Chung LinApr 15 at 17:00

You Can prove by so many methods that $$\forall s\ge0, \log(1+s)\le s \tag{1}$$
Now use $$(1)$$ with $$s={n\over t}$$ and take exponential of both side
$$\log\big(1+\frac{n}{t}\big)\le \frac{n}{t}\iff t\log\big(1+\frac{n}{t}\big)\le n\iff \log\big(1+\frac{n}{t}\big)^t\le n\iff\left(1+\frac{n}{t}\right)^t\ \le\ e^n$$