Simply connected reduced suspension on path connected X

If X is path connected how may i show that the reduced Suspension $\Sigma$ X is then simply connected. I cannot seem to picture this construction

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Do you know the seifert can kampen theorem? – mland Sep 13 '12 at 7:01
Whilst I thought it was true immediately by the Freudenthal suspension theorem, there are a serious of posts by Tom Goodwillie lehigh.edu/~dmd1/tg26, lehigh.edu/~dmd1/tg27, lehigh.edu/~dmd1/tg28 that suggests it might not (always be true). Although I can't see any flaw in the argument provided below... – Juan S Sep 13 '12 at 11:38
@JuanS: I haven't read the posts in detail, but in the first one the counterexample he produces involved the Hawaiian earring which does not satisfy the hypotheses of Freudenthal's suspension theorem. – lentic catachresis Mar 10 '14 at 12:31

An overkilling answer could be the following: The Freudenthal suspension theorem tells us that, if $X$ is $n$-connected, then the natural morphism
$$\pi_k(X) \longrightarrow \pi_{k+1}(\Sigma X)$$
is an isomorphism for $k\leq 2n$. Particularly, for $n=0$, we have an isomorphism $\pi_1(\Sigma X) = 1$.
But, if you want to "picture" the situation, take a look at this suspension drawing, and use the Seifert-van Kampen theorem, as mland points you. Particularly, look at Wikipedia's computation of $\pi_1(S^2)$.
Well, this doesn't say why $\pi_0(\Sigma X)=0$ :) (which can be seen just from the definition of $\Sigma X$). – lentic catachresis Jan 14 '15 at 16:22