prove by induction $(1+a)^{1/n} \leq 1+a/n$ while $a\geq-1$

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So, I've tried to solve this by induction, but without success. I get this equation: $(1+a)^{1/k}\leq 1+a/k$ and this equation, that I have to prove: $(1+a)^{1/(k+1)}\leq1+a/(k+1)$

I tried numerous ways, but without success, the last thing that I tried was to isolate the $1$ on both equations, $(1+a)^{1/k}-a/k\leq1$ $(1+a)^{1/(k+1)}-a/(k+1)\leq1$ and then to try and prove that: $(1+a)^{1/(k+1)}-a/(k+1)<(1+a)^{1/k}-a/k\leq1$

But it didn't work very well.

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$$(1+a)^{\frac{1}{n}}\leq 1+\frac{a}{n}\iff 1+a\leq (1+\frac{a}{n})^n$$

Call $x=\frac{a}{n}$, then we want to prove $$(1+x)^n\geq 1+nx\text{ for } x>-\frac{1}{n}.$$

Let's prove it by induction for $x\geq-1$. $n=1$ is obvious.

$n\Rightarrow n+1$. $$(1+x)^{n+1}=(1+x)(1+x)^n\geq(1+x)(1+nx)=1+(n+1)x+nx^2\geq 1+(n+1)x$$